LIBRARY 

OF   THE 

UNIVERSITY  OF  CALIFORNIA. 
Deceived          WAR  11   1893.  ,89 

Accessions  No. ^^  .  Class  No. 

Engineering 

Library 


APPLIED   MECHANICS 


BY 

GAETANO   LANZA,   S.B.,  C.&M.E., 

PROFESSOR  OF  THEORETICAL  AND   APPLIED   MECHANICS,   MASSACHUSETTS 
INSTITUTE  OF  TECHNOLOGY. 


SIXTH  EDITION,  REVISED  AND  ENLARGED, 


NEW    YORK: 
JOHN     WILEY    &    SONS, 

53  EAST  TENTH  STREET. 
1892. 


T>4 


Engineering 
Library 


COPYRIGHT,  1885, 
BY  GAETANO   LANZA. 


PREFACE. 


THIS  book  is  the  result  of  the  experience  of  the  writer 
in  teaching  the  subject  of  Applied  Mechanics  for  the  last 
twelve  years  at  the  Massachusetts  Institute  of  Technology. 

The  immediate  object  of  publishing  it  is,  to  enable  him  to 
dispense  with  giving  to  the  students  a  large  amount  of  notes. 
As,  however,  it  is  believed  that  it  may  be  found  useful  by 
others,  the  following  remarks  in  regard  to  its  general  plan 
are  submitted. 

The  work  is  essentially  a  treatise  on  strength  and  stabil- 
ity; but,  inasmuch  as  it  contains  some  other  matter,  it  was 
thought  best  to  call  it  "  Applied  Mechanics,"  notwithstanding 
the  fact  that  a  number  of  subjects  usually  included  in  trea- 
tises on  applied  mechanics  are  omitted. 

It  is  primarily  a  text-book ;  and  hence  the  writer  has  endeav- 
ored to  present  the  different  subjects  in  such  a  way  as 
seemed  to  him  best  for  the  progress  of  the  class,  even  though 
it  be  at  some  sacrifice  of  a  logical  order  of  topics.  While 
no  attempt  has  been  made  at  originality,  it  is  believed  that 
some  features  of  the  work  are  quite  different  from  all  pre- 


IV  PREFACE. 


vious  efforts ;  and  a  few  of  these  cases  will  be  referred  to, 
with  the  reasons  for  so  treating  them. 

In  the  discussion  upon  the  definition  of  "force,"  the  object 
is,  to  make  plain  to  the  student  the  modern  objections  to  the 
usual  ways  of  treating  the  subject,  so  that  he  may  have  a 
clear  conception  of  the  modern  aspect  of  the  question,  rather 
than  to  support  the  author's  definition,  as  he  is  fully  aware 
that  this,  as  well  as  all  others  that  have  been  given,  is  open 
to  objection. 

In  connection  with  the  treatment  of  statical  couples,  it 
was  thought  best  to  present  to  the  student  the  actual  effect 
of  the  action  of  forces  on  a  rigid  body,  and  not  to  delay  this 
subject  until  dynamics  of  rigid  bodies  is  treated,  as  is  usually 
done. 

In  the  common  theory  of  beams,  the  author  has  tried  to 
make  plain  the  assumptions  on  which  it  is  based.  A  little 
more  prominence  than  usual  has  also  been  given  to  the  longi- 
tudinal shearing  of  beams. 

In  that  part  of  the  book  that  relates  to  the  experimental 
results  on  strength  and  elasticity,  the  writer  has  endeavored 
to  give  the  most  reliable  results,  and  to  emphasize  the  fact, 
that,  to  obtain  constants  suitable  for  use  in  practice,  we 
must  deduce  them  from  tests  on  full-size  pieces.  This  prin- 
ciple of  being  careful  not  to  apply  experimental  results  to 
cases  very  different  from  those  experimented  upon,  has  long 
been  recognized  in  physics,  and  therefore  needs  no  justifica- 
tion. 

The  government  reports  of  tests  made  at  the  Watertown 
Arsenal  have  been  extensively  quoted  from,  as  it  is  believed 


PREFACE. 


that   they  furnish  some  of   our  most  reliable  information  on 
these  subjects. 

The  treatment  of  the  strength  of  timber  will  be  found  to 
be  quite  different  from  what  is  usually  given;  but  it  speaks 
for  itself,  and  will  not  be  commented  upon  here. 

In  the  chapter  on  the  "  Theory  of  Elasticity,"  a  combina- 
tion is  made  of  the  methods  of  Rankine  and  of  Grashof. 

In  preparing  the  work,  the  author  has  naturally  consulted 
the  greater  part  of  the  usual  literature  on  these  subjects ;  and, 
whenever  he  has  drawn  from  other  books,  he  has  endeavored  to 
acknowledge  it.  He  wishes  here  to  acknowledge  the  assist- 
ance furnished  him  by  Professor  C.  H.  Peabody  of  the  Massa- 
chusetts Institute  of  Technology,  who  has  read  all  the  proofs, 
and  has  aided  him  materially  in  other  ways  in  getting  out  the 
work. 

GAETANO  LANZA. 

MASSACHUSETTS  INSTITUTE  OF  TECHNOLOGY, 

April,  1885. 


PREFACE  TO  THE  FOURTH  EDITION. 

THE  principal  differences  between  this  a'nd  the  earlier 
editions  consist  in  the  introduction  of  the  results  of  a  large 
amount  of  the  experimental  work  that  has  been  done  during 
the  last  five  years  upon  the  strength  of  materials. 

The  other  changes  that  have  been  made  in  the  book  are  not 
a  great  many,  and  have  been  suggested  as  desirable  by  the 
author's  experience  in  teaching. 

September,  1890. 


TABLE  OF  CONTENTS. 


CHAPTER  I. 

COMPOSITION  AND  RESOLUTION  OF  FORCES  .    , 


CHAPTER  II. 
DYNAMICS 75 

CHAPTER  III. 
ROOF-TRUSSES 138 

CHAPTER  IV. 
BRIDGE-TRUSSES    . 184 

CHAPTER  V. 
CENTRE  OF  GRAVITY 221 

CHAPTER  VI. 
STRENGTH  OF  MATERIALS 240 

CHAPTER  VII. 
STRENGTH  OF  MATERIALS  AS  DETERMINED  BY  EXPERIMENT 350 


Vlll  TABLE   OF  CONTENTS. 

CHAPTER  VIII. 
CONTINUOUS  GIRDERS 743 

CHAPTER  IX. 
EQUILIBRIUM  CURVES. —  ARCHES  AND  DOMES 779 

CHAPTER  X. 
THEORY  OF  ELASTICITY,  AND  APPLICATIONS 852 


APPLIED- "MECHANICS. 


CHAPTER   I. 

COMPOSITION  AND  RESOLUTION  OF  FORCES. 

§  i.  Fundamental  Conceptions.  —  The  fundamental  con- 
ceptions of  Mechanics  are  Force,  Matter,  Space,  Time,  and 
Motion. 

§  2.  Relativity  of  Motion.  —  The  limitations  of  our  natures 
are  such  that  all  our  quantitative  conceptions  are  relative. 
The  truth  of  this  statement  may  be  illustrated,  in  the  case  of 
motion,  by  the  fact,  that,  if  we  assume  the  shore  as  fixed  in 
position,  a  ship  sailing  on  the  ocean  is  in  motion,  and  a  ship 
moored  in  the  dock  is  at  rest ;  whereas,  if  we  assume  the  sun 
as  our  fixed  point,  both  ships  are  really  in  motion,  as  both  par- 
take of  the  motion  of  the  earth.  We  have,  moreover,  no  means 
of  determining  whether  any  given  point  is  absolutely  fixed  in 
position,  nor  whether  any  given  direction  is  an  absolutely  fixed 
direction.  Our  only  way  of  determining  direction  is  by  means 
of  two  points  assumed  as  fixed  ;  and  the  straight  line  joining 
them,  we  are  accustomed  to  assume  as  fixed  in  direction. 
Thus,  it  is  very  customary  to  assume  the  straight  line  joining 
the  sun  with  any  fixed  star  as  a  line  fixed  in  direction ;  but  if 
the  whole  visible  universe  were  in  motion,  so  as  to  change  the 
absolute  direction  of  this  line,  we  should  have  no  means  of 
recognizing  it. 


APPLIED  MECHANICS. 


§3.  Rest  and  Motion.  —  In  order  to  define  rest  and 
motion,  we  have  the  following  ;  viz.,  — 

When  a  single  point  is  spoken  of  as  having  motion  or  rest, 
some  other  point  is  always  expressed  or  understood,  which  is 
for  the  time  being  considered  as  a  fixed  point,  and  some  direc- 
tion is  assumed  as  a  fixed  direction  :  and  we  then  say  that  the 
first-named  point  is  at  rest  relatively  to  the  fixed  point,  when 
the  straight  line  joining  it  with  the  fixed  point  changes  neither 
in  length,  nor  in  direction;  whereas  it  is  said  to  be  in  motion 
relatively  to  the  fixed  point,  when  this  straight  line  changes  in 
length,  in  direction,  or  in  both. 

If,  on  the  other  hand,  we  had  considered  the  first-named 
point  as  our  fixed  point,  the  same  conditions  would  determine 
whether  the  second  was  at  rest,  or  in  motion,  relatively  to  the 
first. 

A  body  is  said  to  be  at  rest  relatively  to  a  given  point  and 
to  a  given  direction,  when  all  its  points  are  at  rest  relatively  to 
this  point  and  this  direction. 

§  4.  Velocity.  —  When  the  motion  of  one  point  relatively 
to  another,  or  of  one  body  relatively  to  another,  is  such  that  it 
describes  equal  distances  in  equal  times,  however  small  be  the 
parts  into  which  the  time  is  divided,  the  motion  is  said  to  be 
uniform  and  the  velocity  constant. 

The  velocity,  in  this  case,  is  the  space  passed  over  in  a  unit 
of  time,  and  is  to  be  found  by  dividing  the  space  passed  over  in 
any  given  time  by  the  time ;  thus,  if  s  represent  the  space 
passed  over  in  time  /,  and  v  represent  the  velocity,  we  shall 
have 


When  the  motion  is  not  uniform,  if  we  divide  the  time  into 
small  parts,  and  then  divide  the  space  passed  over  in  one  of 
these  intervals  by  the  time,  and  then  pass  to  the  limit  as  these 
intervals  of  time  become  shorter,  we  shall  obtain  the  velocity. 


FORCE.  3 


Thus,  if  AJ  represent  the  space  passed  over  in  the  interval  of 
time  A^,  then  we  shall  have 

V  =  limit  of  —  as  A/  diminishes, 
A/ 

or 


§  5.  Force  --  We  shall  next  attempt  to  obtain  a  correct  defi- 
nition of  force,  or  at  least  of  what  is  called  force  in  mechanics. 

It  may  seem  strange  that  it  should  be  necessary  to  do  this  ; 
as  it  would  appear  that  clear  and  correct  definitions  must  have 
been  necessary  in  order  to  make  correct  deductions,  and  there- 
fore that  there  ought  to  be  no  dispute  whatever  over  the  mean- 
ing of  the  word  force.  Nevertheless,  it  is  a  fact  in  mechanics, 
as  well  as  in  all  those  sciences  which  attempt  to  deal  with  the 
facts  and  laws  of  nature,  that  correct  definitions  are  only  gradu- 
ally developed,  and  that,  starting  with  very  imperfect  and  often 
erroneous  views  of  natural  laws  and  phenomena,  it  is  only  after 
these  errors  have  been  ascertained  and  corrected  by  a  long 
range  of  observation  and  experiment,  and  an  increased  range  of 
knowledge  has  been  acquired,  that  exactness  and  perspicuity 
can  be  obtained  in  the  definitions. 

Now,  this  is  precisely  what  has  happened  in  the  case  of 
force. 

In  ancient  times  rest  was  supposed  to  be  the  natural  state 
of  bodies  ;  and  it  was  assumed  that,  in  order  to  make  them 
move,  force  was  necessary,  and  that  even  after  they  had  been 
set  in  motion  their  own  innate  inertia  or  sluggishness  would 
cause  them  to  come  to  rest  unless  they  were  constantly  urged 
on  by  the  application  of  some  force,  the  bodies  coming  to  rest 
whenever  the  force  ceased  acting. 

It  was  under  the  influence  of  these  vague  notions  that  such 
terms  arose  as  Force,  of  Inertia,  Moment  of  Inertia,  Vis^  Viva 
or  Living  Force,  etc. 


APPLIED   MECHANICS. 


A  number  of  these  terms  are  still  used  in  mechanics  ;  but  in 
all  such  cases  they  have  been  re-defined,  such  new  meanings 
having  been  attached  to  them  as  will  bring  them  into  accord 
with  the  more  advanced  ideas  of  the  present  time.  Such  defini- 
tions will  be  given  in  the  course  of  this  work,  as  the  necessity 
may  arise  for  the  use  of  the  terms. 

Moreover,  it  is  to  be  regretted  that  there  still  prevails,  to 
some  extent,  in  the  more  popular  class  of  scientific  literature,  a 
loose  usage  of  such  terms,  which  is  very  liable  to  impart  erro- 
neous ideas  to  those  whose  minds  are  not  clear  as  to  their  true 
meanings. 

NEWTON'S    FIRST    LAW    OF    MOTION. 

Ideas  becoming  more  precise,  in  course  of  time  there  was 
framed  Newton's  first  law  of  motion  ;  and  this  law  is  as  fol- 
lows :  — 

A  body  at  rest  will  remain  at  rest,  and  a  body  in  motion  will 
continue  to  move  uniformly  and  in  a  straight  line,  unless  and 
until  some  external  force  acts  upon  it. 

The  assumed  truth  of  this  law  was  based  upon  the  observed 
facts  of  nature  ;  viz.,  — 

When  bodies  were  seen  to  be  at  rest,  and  from  rest  passed 
into  a  state  of  motion,  it  was  always  possible  to  assign  some 
cause ;  i.e.,  they  had  been  brought  into  some  new  relationship, 
either  with  the  earth,  or  with  some  other  body:  and  to  this 
cause  could  be  assigned  the  change  of  state  from  rest  to  motion. 
On  the  other  hand,  in  the  case  of  bodies  in  motiop,  it  was  seen, 
that,  if  a  body  altered  its  motion  from  a  uniform  rectilinear 
motion,  there  was  always  some  such  cause  that  could  be 
assigned.  Thus,  in  the  case  of  a  ball  thrown  from  the  hand, 
the  attraction  of  the  earth  and  the  resistance  of  the  air  soon 
caused  it  to  come  to  rest.  In  the  case  of  a  ball  rolled  along 
the  ground,  friction  (i.e.,  the  continual  contact  and  collision  with 
the  ground)  gradually  destroyed  its  motion,  and  brought  it  to 


FORCE. 


rest ;  whereas,  when  such  resistances  were  diminished  by  rolling 
it  on  glass  or  on  the  ice,  the  motion  always  continued  longer : 
hence  it  was  inferred,  that,  were  these  resistances  entirely 
removed,  the  motion  would  continue  forever.  » 

In  accordance  with  these  views,  the  definition  of  force 
usually  given  was  substantially  as  follows  :  — 

Force  is  that  which  causes,  or  tends  to  cause,  a  body  to  change 
its  state  from  rest  to  motion,  from  motion  to  rest,  or  to  change  its 
motion  as  to  direction  or  speed. 

Under  these  views,  uniform  rectilinear  motion  was  recog- 
nized as  being  just  as  much  a  condition  of  equilibrium,  or  of 
the  action  of  no  force  or  of  balanced  forces,  as  rest ;  and  the 
recognition  of  this  one  fact  upset  many  false  notions,  destroyed 
many  incorrect  conclusions,  and  first  rendered  possible  a  science 
of  mechanics.  Along  with  the  above-stated  definition  of  force 
is  ordinarily  given  the  following  proposition  ;  viz.,  — 

Forces  are  proportional  to  the  velocities  that  they  impart,  in  a 
unit  of  time,  to  tJie  same  body.  The  reasoning  given  in  support 
of  .this  proposition  is  as  follows  :  — 

Suppose  a  body  to  be  moving  uniformly  and  in  a  straight 
line,  and  suppose  a  force  to  act  upon  it  for  a  certain  length  of 
time  /  in  the  direction  of  the  body's  motion  :  the  effect  of  the 
force  is  to  alter  the  velocity  of  the  body ;  and  it  is  only  by  this 
alteration  of  velocity  that  we  recognize  the  action  of  the  force. 
Hence,  as  long  as  the  alteration  continues  at  the  same  rate,  we 
recognize  the  same  force  as  acting. 

If,  therefore,/  represent  the  amount  of  velocity  which  the 
force  would  impart  in  one  unit  of  time,  the  total  increase  in 
the  velocity  of  the  body  will  be  ft;  and,  if  the  force  now  stop 
acting,  the  body  will  again  move  uniformly  and  in  the  same 
direction,  but  with  a  velocity  greater  by//. 

Hence,  if  we  are  to  measure  forces  by  their  effects,  it  will 
follow  that  — 

The  velocity  which  a  force  will  impart  to  a  given  (or  standard) 


APPLIED  MECHANICS. 


body  in  a  unit  of  time  is  a  proper  measure  of  the  force.  And 
we  shall  have,  that  two  forces,  each  of  which  will  impart  the 
same  velocity  to  the  same  body  in  a  unit  of  time,  are  equal  to 
each  other  ;  and  a  force  which  will  impart  to  a  given  body  twice 
the  velocity  per  unit  of  time  that  another  force  will  impart  to 
the  same  body,  is  itself  twice  as  great,  or,  in  other  words,  — 

Forces  are  proportional  to  the  velocities  that  they  impart,  in  a 
unit  of  time,  to  the  same  body. 

MODERN   CRITICISM   OF    THE   ABOVE. 

The  scientists  and  the  metaphysicians  of  the  present  time 
are  recognizing  two  other  facts  not  hitherto  recognized,  and  the 
result  is  a  criticism  adverse  to  the  above-stated  definition  of 
force.  Other  definitions  have,  in  consequence,  been  proposed; 
but  none  are  free  from  objection  on  logical  grounds,  and  at  the 
same  time  capable  of  use  in  mechanics  in  a  quantitative  way. 

The  two  facts  referred  to  are  the  following ;  viz.,  — 

i°.  That  all  our  ideas  of  space,  time,  rest,  motion,  and  even 
of  direction,  are  relative. 

2°.  That,  because  two  effects  are  identical,  it  does  not  follow 
that  the  causes  producing  those  effects  are  identical. 

Hence,  in  the  light  of  these  two  facts,  it  is  plain,  that,  inas- 
much as  we  can  only  recognize  motion  as  relative,  we  can  only 
recognize  force  as  acting  when  at  least  two  bodies  are  con- 
cerned in  the  transaction  ;  and  also  that  if  the  forces  are  simply 
the  causes  of  the  motion  in  the  ordinary  popular  sense  of  the 
word  cause,  we  cannot  assume,  that,  when  the  effects  are  equal, 
the  causes  are  in  every  way  identical,  although  we  have,  of 
course,  a  perfect  right  to  say  that  they  are  identical  so  far  a3 
the  production  of  motion  is  concerned. 

I  shall  now  proceed,  in  the  light  of  the  above,  to  deduce  a 
definition  of  force,  which,  although  not  free  from  objection, 
seems  as  free  as  any  that  has  been  framed. 

It  is  one  of  the  facts  of  nature,  that,  when  bodies  are  by  any 


FORCE. 


means  brought  under  certain  relations  to  each  other,  certain 
tendencies  are  developed,  which,  if  not  interfered  with,  will 
exhibit  themselves  in  the  occurrence  of  certain  definite  phe- 
nomena. What  these  phenomena  are,  depends  upon  the  nature 
of  the  bodies  concerned,  and  on  the  relationships  into  which 
they  are  brought. 

As  an  illustration,  we  know  that  if  an  apple  is  placed  at  a 
certain  height  above  the  surface  of  the  earth,  there  is  developed 
between  the  two  bodies  a  tendency  to  approach  each  other ; 
and  if  there  is  no  interference  with  this  tendency,  it  exhibits 
itself  in  the  fall  of  the  apple.  If,  on  the  other  hand,  the  apple 
were  hung  on  the  hook  of  a  spring  balance  in  the  same  posi- 
tion as  before,  the  spring  would  stretch,  and  there  would  be 
developed  a  tendency  of  the  spring  to  make  the  apple  move 
upwards.  This  tendency  to  make  the  apple  move  upwards 
would  be  just  equal  to  the  tendency  of  the  earth  and  apple  to 
approach  each  other.  This  would  be  expressed  by  saying  that 
the  pull  of  the  spring  is  just  equal  and  opposite  to  the  weight 
of  the  apple. 

As  other  illustrations  of  these  tendencies  developed  in 
bodies  when  placed  in  certain  relations  to  each  other,  we  have 
the  following  cases  :  — 

(a)  When  two  bodies  collide. 

(b)  When  two  substances,  coming  together,  form  a  chemical 
union,  as  sodium  and  water. 

(c)  When  the  chemical  union  is  entered  into  only  by  raising 
the  temperature  to  some  special  point. 

Any  of  these  tendencies  that  are  developed  by  bringing 
about  any  of  these  special  relationships  between  bodies  might 
properly  be  called  a  force  ;  and  the  term  might  properly  be,  and 
is,  used  in  the  same  sense  in  the  mental  and  moral  world,  as 
well  as  in  the  physical.  In  mechanics,  however,  we  have  to 
.deal  only  with  the  relative  motion  of  bodies;  and  hence  we 
give  the  name  force  only  to  tendencies  to  change  the  relative 


8  APPLIED   MECHANICS. 

motion  of  the  bodies  concerned ;  and  this,  whether  these  ten- 
dencies are  unresisted,  and  exhibit  themselves  in  the  actual 
occurrence  of  a  change  of  motion,  or  whether  they  are  resisted 
by  equal  and  opposite  tendencies,  and  exhibit  themselves  in 
the  production  of  a  tensile,  compressive,  or  other  stress  in  the 
bodies  concerned,  instead  of  motiorf. 

DEFINITION    OF    FORCE. 

Hence  our  definition  of  force,  as  far  as  mechanics  has  to 
deal  with  it  or  is  capable  of  dealing  with  it,  is  as  follows; 
viz.,— 

Force  is  a  tendency  to  change  the  relative  motion  of  the  two 
bodies  between  which  that  tendency  exists. 

Indeed,  when,  as  in  the  illustration  given  a  short  time  ago, 
the  apple  is  hung  on  the  hook  of  a  spring  balance,  there  still 
exists  a  tendency  of  the  apple  and  the  earth  to  approach  each 
other  ;  i.e.,  they  are  in  the  act  of  trying  to  approach  each  other ; 
and  it  is  this  tendency,  or  act  of  trying,  that  we  call  the  force  of 
gravitation.  In  the  case  cited,  this  tendency  is  balanced  by 
an  opposite  tendency  on  the  part  of  the  spring ;  but,  were  the 
spring  not  there,  the  force  of  gravitation  would  cause  the  apple 
to  fall. 

Professor  Rankine  calls  force  "an  action  between  two  bodies, 
either  causing  or  tending  to  cause  change  in  their  relative  rest 
or  motion ;"  and  if  the  act  of  trying  can  be  called  an  action,  my 
definition  is  equivalent  to  his. 

For  the  benefit  of  any  one  who  wishes  to  follow  out  the 
discussions  that  have  lately  taken  place,  I  will  enumerate  the 
following  articles  that  have  been  written  on  the  subject :  — 

(a)  "Recent  Advances  in  Physical  Science,"  by  P.  G.  Tait, 
Lecture  XIV. 

(b)  Herbert    Spencer,    "  First    Principles    of    Philosophy "  . 
(certain  portions  of  the  book). 


MEASURE   OF  FORCE.  9 

(c)  Discussion  by  Messrs.  Spencer  and  Tait,  "  Nature,"  Jan. 
2,  9,  1 6,  1879. 

(d)  Force  and  Energy,  "Nature,"  Nov.  25,  Dec.  2,  9,   16, 
1880. 

§6.  External  Force.  —  We  thus  see,  that,  in  order  that  a 
force  may  be  developed,  there  must  be  two  bodies  concerned 
in  the  transaction ;  and  we  should  speak  of  the  force  as  that 
developed  or  existing  between  the  two  bodies. 

But  we  may  confine  our  attention  wholly  to  the  motion  or 
condition  of  one  of  these  two  bodies ;  and  we  may  refer  its 
motion  either  to  the  other  body  as  a  fixed  point,  or  to  some 
body  different  from  either ;  and  then,  in  speaking  of  the  force, 
we  should  speak  of  it  as  the  force  acting  on  the  body  under 
consideration,  and  call  it  an  external  force.  It  is  the  tendency 
of  the  other  body  to  change  the  motion  of  the  body  under  con- 
sideration relatively  to  the  point  considered  as  fixed. 

§  7.  Relativity  of  Force.  —  In  adopting  the  above-stated 
definition  of  force,  we  acknowledge  our  incapacity  to  deal  with 
it  as  an  absolute  quantity ;  for  we  have  defined  it  as  a  tendency 
to  change  the  relative  motion  of  a  pair  of  bodies.  Hence  it  is 
only  through  relative  motion  that  we  recognize  force;  and  hence 
force  is  relative,  as  well  as  motion. 

§8.  Newton's  First  Law  of  Motion.  —  In  the  light  of 
the  above  discussion,  we  might  express  Newton's  first  law  of 
motion  as  follows  :  — 

A  body  at  rest,  or  in  uniform  rectilinear  motion  relatively  to 
a  given  point  assumed  as  fixed,  will  continue  at  rest,  or  in  tmi~ 
form  motion  in  the  same  direction,  unless  and  until  some  external 
force  acts  either  on  the  body  in  question,  or  on  the  fixed  point, 
or  on  the  body  which  furnishes  us  our  fixed  direction.  This  law 
is  really  superfluous,  as  it  has  all  been  embodied  in  the  defini- 
tion. 

§9.  Measure  of  Force.  —  We  next  need  some  means  of 
comparing  forces  with  each  other  in  magnitude;  and,  subse- 


10  APPLIED  MECHANICS. 

quently,  we  need  to  select  one  force  as  our  unit  force,  by  means 
of  which  to  estimate  the  magnitude  of  other  forces. 

Let  us  suppose  a  body  moving  uniformly  and  in  a  straight 
line,  relatively  to  some  fixed  point ;  as  long  as  this  motion 
continues,  we  recognize  no  unbalanced  force  acting  on  it ; 
but,  if  the  motion  changes,  there  must  be  a  tendency  to  change 
that  motion,  or,  in  other  words,  an  unbalanced  force  is  acting 
on  the  body  from  the  instant  when  it  begins  to  change  its 
motion. 

Suppose  a  body  to  be  moving  uniformly,  and  a  force  to  be 
applied  to  it,  and  to  act  for  a  length  of  time  /,  and  to  be  so  applied 
as  not  to  change  the  direction  of  motion  of  the  body,  but  to 
increase  its  velocity;  the  result  will  be,  that  the  velocity  will  be 
increased  by  equal  amounts  in  equal  times,  and  if  f  represent 
the  amount  of  velocity  the  force  would  impart  in  one  unit  of 
time,  the  total  increase  in  velocity  will  be  ft.  This  results 
merely  from  the  definition  of  a  force ;  for  if  the  velocity  pro- 
duced in  one  (a  standard)  body  by  a  given  force  is  twice  as 
great  as  that  produced  by  another  given  force,  then  is  the  ten- 
dency to  produce  velocity  twice  as  great  in  the  first  case  as  in 
the  second,  or,  in  other  words,  the  first  force  is  twice  as  great 
as  the  second.  Hence  — 

Forces  are  proportional  to  the  velocities  which  they  will  impart 
to  a  given  (or  standard}  body  in  a  unit  of  time. 

We  may  thus,  by  using  one  standard  body,  determine  a 
set  of  equal  forces,  and  also  the  proportion  between  different 
forces. 

§10.  Measure  of  Mass.  —  After  having  determined,  as 
shown,  a  set  of  equal  (unit)  forces,  if  we  apply  two  of  them 
to  different  bodies,  and  let  them  act  for  the  same  length  of  time 
on  each,  and  find  that  the  resulting  velocities  are  unequal,  these 
bodies  are  said  to  have  unequal  masses ;  whereas,  if  the  result- 
ing velocities  are  equal,  they  are  said  to  have  equal  masses. 

Hence  we  have  the  following  definitions  :  — 


RELATION  BETWEEN  FORCE  AND   MOMENTUM.  II 

I  °.  Equal  forces  are  those  which,  by  acting  for  equal  times 
on  the  same  or  standard  body,  impart  to  it  equal  velocities. 

2°.  Equal  masses  are  those  masses  to  which  equal  forces 
will  impart  equal  velocities  in  eqzial  times. 

§11.  Suppose  two  bodies  of  equal  mass  moving  side  by 
side  with  the  same  velocity,  and  uniformly,  let  us  apply  to 
one  of  them  a  force  F  in  the  direction  of  the  body's  motion  : 
the  effect  of  this  force  is  to  increase  the  velocity  with  which  the 
body  moves ;  and  if  we  wish,  at  the  same  time,  to  increase 
the  velocity  of  the  other,  so  that  they  will  continue  to  move 
side  by  side,  it  will  be  necessary  to  apply  an  equal  force  to  that 
also. 

We  are  thus  employing  a  force  2F  to  impart  to  the  two 
bodies  the  required  increment  of  velocity. 

If  we  unite  them  into  one,  it  still  requires  a  force  2F  tp 
impart  to  the  one  body  resulting  from  their  union  the  re- 
quired increment  of  velocity :  hence,  if  we  double  the  mass 
to  which  we  wish  to  impart  a  certain  velocity,  we  must  double* 
the  force,  or,  in  other  words,  employ  a  force  which  would 
impart  to  the  first  mass  alone  a  velocity  double  that  required. 
Hence  — 

Forces  are  proportional  to  the  masses  to  which  they  will  impart 
tJie  same  velocity  in  the  same  time.  . 

§  12.  Momentum.  —  The  product  obtained  by  multiplying 
the  number  of  units  of  mass  in  a  body  by  its  velocity  is  called 
the  momentum  of  the  body. 

§13.  Relation  between  Force  and  Momentum.  —  The 
number  of  units  of  momentum  imparted  to  a  body  in  a  unit  of 
time  by  a  given  force,  is  evidently  identical  with  the  number 
of  units  of  velocity  that  would  be  imparted  by  the  same  force, 
in  the  same  time,  to  a  unit  mass.  Hence  — 

Forces  are  proportional  to  the  momenta  (or  velocities  per  unit 
of  mass]  which  they  will  generate  in  a  unit  of  time. 


12  APPLIED  MECHANICS. 

Hence,  if  F  represent  a  force  which  generates,  in  a  unit  of 
time,  a  velocity/  in  a  body  whose  mass  is  m,  we  shall  have 


and,  inasmuch  as  the  choice  of  our  units  is  still  under  our  con- 
trol, we  so  choose  them  that 

F  =  mf; 

i.e.,  the  force  F  contains  as  many  units  of  force  as  mf  contains 
units  of  momentum  ;  in  other  words,  — 

The  momentum  generated  in  a  body  in  a  unit  of  time  by  a 
force  acting  in  the  direction  of  the  body  s  motion,  is  taken  as 
a  measure  of  the  force. 

§  14.  Statical  Measure  of  Force.  —  When  the  forces  are 
prevented  from  producing  motion  by  being  resisted  by  equal 
and  opposite  forces,  as  is  the  case  in  that  part  of  mechanics 
known  as  Statics,  the*y  must  be  measured  by  a  direct  comparison 
with  other  forces.  An  illustration  of  this  has  already  been 
«given  in  the  case  of  an  apple  hung  on  the  hook  of  a  spring 
balance.  In  that  case  the  pull  of  the  spring  is  equal  in  magni- 
tude to  the  weight  of  the  apple:  indeed,  it  is  very  customary 
to  adopt  for  forces  what  is  known  as  the  gravity  measure,  in 
which  case  we  take  as  our  unit  the  gravitation,  or  tendency  to 
fall,  of  a  given  piece  of  metal,  at  a  given  place  on  the  surface 
of  the  earth  ;  in  other  words,  its  weight  at  a  given  place. 

The  gravity  unit  may  thus  be  the  kilogram,  the  pound,  or 
the  ounce,  etc. 

It  is  evident,  moreover,  from  our  definition  of  force,  and  the 
subsequent  discussion,  that  whatever  we  take  as  our  unit  of 
mass,  the  statical  measure  of  a  force  is  proportional  to  its 
dynamical  measure  ;  i.e.,  the  numbers  representing  the  magni- 
tudes of  any  two  forces,  in  pounds,  are  proportional  to  the 
momenta  they  will  impart  to  any  body  in  a  unit  of  time. 

§15.  Gravity  Measure  of  Mass.  —  If  we  assume  one 
pound  as  our  unit  of  force,  one  foot  as  our  unit  of  length,  and 


NEWTON'S  SECOND  LAW  OF  MOTION  13 

one  second  as  our  unit  of  time,  the  ratio  between  the  number 
of  pounds  in  any  given  force  and  the  momentum  it  will  impart 
to  a  body  on  which  it  acts  unresisted  for  a  unit  of  time,  will 
depend  on  our  unit  of  mass  ;  and,  as  we  are  still  at  liberty  to  fix 
this  as  we  please,  it  will  be  most  convenient  so  to  choose  it 
that  the  above-stated  ratio  shall  be  unity,  so  that  there, shall  be 
no  difference  in  the  measure  of  a  force,  whether  it  is  measured 
statically  or  dynamically.  Now,  it  is  known  that  a  body  falling 
freely  under  the  action  of  its  own  weight  acquires,  every  second, 
a  velocity  of  about  thirty-two  feet  per  second  :  this  number  is 
denoted  by  g,  and  varies  for  different  distances  from  the  centre 
of  the  earth,  as  does  also  the  weight  of  the  body. 

Now,  if  W  represent  the  weight  of  the  body  in  pounds,  and 
m  the  number  of  units  of  mass  in  its  mass,  we  must  have,  in 
order  that  the  statical  and  dynamical  measures  may  be  equal, 

W=  mg. 
Hence 

*»»; 

g 

i.e.,  the  number  of  units  of  mass  in  a  body  is  obtained  by  divid- 
ing the  weight  in  pounds,  by  the  value  of  g  at  the  place  where 
the  weight  is  determined. 

The  values  of  W  and  of  g  vary  for  different  positions,  but 
the  value  of  m  remains  always  the  same  for  the  same  body. 

UNIT    OF    MASS. 

If  m  =  i,  then  W  =  g;  or,  in  words,  — 

The  weight  in  pounds  of  the  unit  of  mass  (when  the  gravity 
measure  is  used}  is  equal  to  the  value  of  g  in  feet  per  second  for 
the  same  place. 

§  16.  Newton's  Second  Law  of  Motion.  —  Newton's 
second  law  of  motion  is  as  follows  :  — 


14  APPLIED   MECHANICS. 

"  Change  of  momentum  is  proportional  to  the  impressed  mov- 
ing force,  and  occurs  along  the  straight  line  in  which  the  force  is 
impressed'' 

Newton  states  further  in  his  "  Principia  :"  — 

"  If  any  force  generate  any  momentum,  a  double  force 
will  generate  a  double,  a  triple  force  will  generate  a  triple, 
momentum,  whether  simultaneously  and  suddenly,  or  gradually 
and  successively  impressed.  And  if  the  body  was  moving 
before,  this  momentum,  if  in  the  same  direction  as  the  motion, 
is  added ;  if  opposite,  is  subtracted ;  or  if  in  an  oblique  direc- 
tion, is  annexed  obliquely,  and  compounded  with  it,  according 
to  the  direction  and  magnitude  of  the  two." 

Part  of  this  law  has  reference  to  the  proportionality  between 
the  force  and  the  momentum  imparted  to  the  body ;  and  this 
has  been  already  embodied  in  our  definition  of  force,  and  illus- 
trated in  the  discussion  on  the  measure  of  forces. 

The  other  part  is  properly  a  law  of  motion,  and  may  be 
expressed  as  follows  :  — 

If  a  body  have  two  or  m»re  velocities  imparted  to  it  simulta- 
neously >  it  will  move  so  as  to  preserve  them  all. 

The  proof  of  this  law  depends  merely  upon  a  proper  con- 
ception of  motion.  To  illustrate  this  law  when  two  velocities 
are  imparted  simultaneously  to  a  body,  let  us  suppose  a  man 
walking  on  the  deck  of  a  moving  ship  :  he  then  has  two  motions 
in  relation  to  the  shore,  his  own  and  that  of  the  ship. 

Suppose  him  to  walk  in  the  direction  of  motion  of  the 
ship  at  the  rate  of  10  feet  per  second;  while  the  ship  moves  at 
25  feet  per  second  relatively  to  the  shore  :  then  his  motion  in 
relation  to  the  shore  will  be  25  +  10  =  35  feet  per  second. 
If,  on  the  other  hand,  he  is  walking  in  the  opposite  direction  at 
the  same  rate,  his  motion  relatively  to  the  shore  will  be  25  — 
10  =  15  feet  per  second. 

Suppose  a  body  situated  at  A  (Fig.  i)  to  have  two  motions 
imparted  to  it  simultaneously,  one  of  which  would  carry  it  to  B 


POLYGON  OF  MOTIONS  15 

in  one  second,  and  the  other  to  C  in  one  second ;  and  that  it  is 
required  to  find  where  it  will  be  at  the  end  of  one  second,  and 
what  path  it  will  have  pursued.  c 

Imagine  the  body  to  move  in  obedience 
to  the  first  alone,  during  one  second  :  it 
would  thus  arrive  at  B ;  then  suppose  the 
second  motion  to  be  imparted  to  the  body, 
instead  of  the  first,  it  will  arrive  at  the  end  of  the  next  sec- 
ond at  D,  where  BD  is  equal  and  parallel  to  AC.  When 
the  two  motions  are  imparted  simultaneously,  instead  of  suc- 
cessively, the  same  point  D  will  be  reached  in  one  second, 
instead  of  two;  and  by  dividing  AB  and  AC  into  the  same 
(any)  number  of  equal  parts,  we  can  prove  that  the  body  will 
.always  be  situated  at  some  point  of  the  diagonal  AD  of  the 
parallelogram,  hence  that  it  moves  along  AD.  Hence  follows 
the  proposition  known  as  the  parallelogram  of  motions. 

PARALLELOGRAM    OF    MOTIONS. 

If  there  be  simultaneously  impressed  on  a  body  tzvo  velocities  y 
which  would  separately  be  represented  by  the  lines  AB  and  AC, 
the  actual  velocity  will  be  represented  by  the  line  AD.  which  is 
the  diagonal  of  the  parallelogram  of  which  AB  and  AC  are  the 
adjacent  sides. 

§17.  Polygon  of  Motions. — In  all  the  above  cases,  the 
point  reached  by  the  body  at  the  end  of  a  second  when  the 
two  motions  take  place  simultaneously  is  the  same  as  that  which 
would  be  reached  at  the  end  of  two  seconds  if  the  motions  took 
place  successively ;  and  the  path  described  is  the  straight  line 
joining  the  initial  position  of  the  body,  with  its  position  at  the 
end  of  one  second  when  the  motions  are  simultaneous. 

The  same  principle  applies  whatever  be  the  number  of 
velocities  that  may  be  imparted  to  a  body  simultaneously. 
Thus,  if  we  suppose  the  several  velocities  imparted  to  be 
(Fig.  2)  AB,  AC,  AD,  AE,  and  AF,  and  it  be  required  to 


l6  APPLIED  MECHANICS. 

determine  the  resultant  velocity,  we  first  let  the  body  move 
with  the  velocity  AB  for  one  second ;  at  the 
end  of  that  second  it  is  found  at  B ;  then  let 
it  move  with  the  velocity  AC  only,  and  at 
the  end  of  another  second  it  will  be  found 
at  c ;  then  with  AD  only,  and  at  the  end  of 
the  third  second  it  will  be  found  at  d;  at  the 
end  of  the  fourth  at  e;  at  the  end  of  the  fifth 
at  f.  Hence  the  resultant  velocity,  when  all 
are  imparted  simultaneously,  is  Aft  or  the 

closing  side  of  the  polygon. 

This  proposition  is  known  as  the  polygon  of  motions. 

POLYGON   OF   MOTIONS. 

If  there  be  simultaneously  impressed  on  a  body  any  number 
of  velocities,  the  resulting  velocity  will  be  represented  by  the 
closing  side  of  a  polygon  of  which  the  lines  representing  tJie 
separate  velocities  form  tJie  other  sides. 

§  1 8.  Characteristics  of  a  Force.  —  A  force  has  three 
characteristics,  which,  when  known,  determine  it ;  viz.,  Point 
of  Application,  Direction,  and  Magnitude.  These  can  be  repre- 
sented by  a  straight  line,  whose  length  is  made  proportional  to 
the  magnitude  of  the  force,  whose  direction  is  that  of  the 
motion  which  the  force  imparts,  or  tends  to  impart,  and  one  end 
of  which  is  the  point  of  application  of  the  force ;  an  arrow-head 
being  usually  employed  to  indicate  the  direction  in  which  the 
force  acts. 

§  19.   Parallelogram  of   Forces. 

PROPOSITION. — If  two  forces  acting  simultaneously  at  the 
same  point  be  represented,  in  point  of  application,  direction, 
and  magnitude,  by  tivo  adjacent  sides  of  a  parallelogram,  their 
resultant  will  be  represented  by  the  diagonal  of  the  parallelo- 
gram, ^/raze/;/ /r<?;#  the  point  of  application  of  the  two  forces. 

PROOF. — In  the  last  part  of  §   16  was  proved  the  propo 


PARALLELOGRAM  OF  FORCES.  \*J 

sition  known  as  the  Parallelogram  of  Motions,  for  the  state- 
ment of  which  the  reader  is  referred  to  the  close  of  that 
section. 

We  have  also  seen  that  forces  are  proportional  to  the  velo- 
cities which  they  impart,  or  tend  to  impart,  in  a  unit  of  time, 
to  the  same  body. 

Hence  the  lines  representing  the  two  impressed  forces  are 
coincident  in  direction  with,  and  proportional  to,  the  lines  repre- 
senting the  velocities  they  would  impart  in  a  unit  of  time  to 
the  same  body  ;  and  moreover,  since  the  resultant  velocity  is 
represented  by  the  diagonal  of  the  parallelogram  drawn  with 
the  component  velocities  as  sides,  the  resultant  force  must  coin- 
cide in  direction  with  the  resultant  velocity,  and  the  length  of 
the  line  representing  the  resultant  force  will  bear  to  the  result 
ant  velocity  the  same  ratio  that  one  of  the  component  forces 
bears  to  the  corresponding  velocity.  Hence  it  follows,  that  the 
resultant  force  will  be  represented  by  the  diagonal  of  the  paral 
lelogram  having  for  sides  the  two  component  forces. 

§  20.   Parallelogram  of   Forces  :  Algebraic  Solution. 

PROBLEM.  —  Given  two  forces  F  and  F,  acting  at  the  same 
point  A  (Fig.  3),  and  inclined  to  each  other  at  an  angle  6;  required 
the  magnitude  and  direction  of  the  resultant 
force. 

Let  AC  represent  F,  AB  represent  Flt 
and  let  angle  BAG  —  0  ;  then  will  R  =  AD    A 
represent  in  magnitude  and  direction   the 
resultant  force.     Also  let  angle  DAC  —  a;  then  from  the  tri- 
angle DAC  we  have 

AD*  =  AC*  +  CD*  -  2AC.  CDcosACD. 
But 

=  -cos0 


.-.     R*       =  F*  +  F*  +  2FFt  cos  0 


-f-  F*  -f  2FF,  cos  6. 


i8 


APPLIED   MECHANICS, 


This  determines  the  magnitude  of  R.  To  determine  its  direc- 
tion, let  angle  CAD  =  a  .'.  angle  BAD  =  0  —  a,  and  we 
shall  have  from  the  triangle  DAC 


or 


and  similarly 


CD  :  AD  =  sin  CAD  :  smACD, 

Ff :  R  =  sin  a  :  sin  0 
sin  a     =  — -sin0, 


sin(<9  -  a)  =      si 


EXAMPLES. 


14'  21";  find  R  and  a. 


i°.  Given  F  =  47-34,  F,  =  75.46,  0  =     73 

2°.  Given  F  =     5.36,  Fl  —  4.27,  0  =     32°  10'  ;  find  R  and  a. 

3°.  Given  F  =  42.00,  7^  =  31.00,  0  =  150°  ;  find  R  and  a. 

4°.  Given  F  =  47.00,  ^  =  75.00,  0  =  253°  ;  find  R  and  a. 

§21.  Parallelogram  of  Forces  when  0=  90°.  —  When 
the  two  given  forces  are  at  right  angles  to  each  other,  the  for- 
mulae become  very  much  simplified,  since  the  parallelogram 
becomes  a  rectangle. 

From  Fig.  4  we  at  once  deduce 


R     = 


COS.  - 


EXAMPLES. 


i°.  Given  F  =     3.0,  Fl  =  5.0  ;  find  R  and  a. 

2°.  Given  F  =     3.0,  Fv  =  —  5.0  ;  find  R  and  a. 

3°.  Given  F  =     5.0,  Fl  =  12.0  ;  find  R  and  a. 

4°.  Given  F  =  23.2,  FL  =  21.3  ;  find  ^  and  a. 


DECOMPOSITION  OF  FORCES  IN  ONE   PLANE.  19 

§  22.  Triangle  of  Forces.  —  If  three  forces  be  represented, 
in  magnitude  and  direction,  by  the  three  sides  of  a  triangle  taken 
in  order,  then,  if  these  forces  be  simultaneously  applied  at  one 
point,  they  will  balance  each  other. 

Conversely,  three  forces  which,  when  simultaneously  applied 
at  one  point,  balance  each  other,  can  be  correctly  represented  in 
magnitude  and  direction  by  the  three  sides  of  a  triangle  taken  in 
order. 

These  propositions,  which  find  a  very  extensive  application, 
especially  in  the  determination  of  the  stresses  in  roof  and 
bridge  trusses,  are  proved  as  follows  :  — 

If  we  have  two  forces,  AC  and  AB  (see  Fig.  3),  acting  at  the 
point  A,  their  resultant  is,  as  we  have  already  seen,  AD ;  and 
hence  a  force  equal  in  magnitude  and  opposite  in  direction  to 
AD  will  balance  the  two  forces  AC  and  AB.  Now,  the  sides  of 
the  triangle  ACDA,  if  taken  in  order,  represent  in  magnitude 
and  direction  the  force  AC,  the  force  CD  or  AB,  and  a  force 
equal  and  opposite  to  AD ;  and  these  three  forces,  if  applied  at 
the  same  point,  would  balance  each  other.  Hence  follows  the 
proposition. 

Moreover,  we  have 

AC  :  CD  :  DA  =  smADC     :  sin  CAD  :  smACZ>, 
or 

F  \  Fl     \  R      =  sin(0  —  a)  :  sin  a          :  sin0; 

or  each  force  is,  in  this  case,  proportional  to  the  sine  of  the 
angle  between  the  other  two. 

§  23.  Decomposition  of  Forces  in  one  Plane.  —  It  is 
often  convenient  to  resolve  a  force  into  two  components,  in  two 
given  directions  in  a  plane  containing  the  force.  Thus,  suppose 
we  have  the  force  R  =  AD  (Fig.  3),  and  we  wish  to  resolve  it 
into  two  components  acting  respectively  in  the  directions  AC 
and  AB ;  i.e.,  we  wish  to  find  two  forces  acting  respectively  in 
these  directions,  of  which  AD  shall  be  the  resultant :  we 


20 


APPLIED   MECHANICS. 


determine  these  components  graphically  by  drawing  a  parallelo- 
gram, of  which  AD  shall  be  the  diagonal,  and  whose  sides  shall 
have  the  directions  AC  and  AB  respectively.  The  algebraic 
values  of  the  magnitudes  of  the  compo- 
nents can  be  determined  by  solving  the 
triangle  ADC.  In  the  case  when  the 
directions  of  the  components  are  at  right 
angles  to  each  other,  let  the  force  R 
(Fig.  5),  applied  at  O,  make  an  angle  a 
with  OX.  We  may,  by  drawing  the  rect- 
angle shown  in  the  figure,  decompose  R 


FIG.  5. 


into  two  components,  F  and  Flt  along  OX  and  O  Y  respectively  ; 
and  we  shall  readily  obtain  from  the  figure, 

F  =  R  cos  a,     Ft  =  R  sin  a. 


EXAMPLES. 

i°.  The  force  exerted  by  the  steam  upon  the  piston  of  a  steam-engine 

at  the  moment  when  it  is  in  the  position  shown  in  the  figure  is  AB  = 

1000   Ibs.      The   resistance   of  the 

guides  upon  the  cross-head  DE  is 

vertical.    Determine  the  force  acting 

along  the   connecting-rod  AC  and 

the    pressure   on   the   guides ;    also 

resolve  the  force  acting  along  the  connecting-rod  into 
two  components,  one   along,  and  the 
other  at  right  angles  to,  the  crank  OC. 
2°.  A  load  of  500  Ibs.  is  placed  at 
the  apex  C  of  the  frame  ACB :   find 


FIG.  6. 


FIG.  7. 


FIG.  8. 


the  stresses  in  AC  and  CB  respectively. 

3°.  A  load  of  4000  Ibs.  is  hung  at  C,  on  -the  crane 
ABC :  find  the  pressure  in  the  boom  BC,  and  the  pull 
on  the  tie  AC,  where  BC  makes  an  angle  of  60°  with  the  horizontal, 
and  AC  an  angle  of  15°. 


COMPOSITION  OF  FORCES  IN  ONE  PLANE.  21 

4°.  A  force  whose  magnitude  is  7  is  resolved  into  two  forces  whose 
magnitudes  are  5  and  3  :  find  the  angles  they  make  with  the  given 
force. 

§  24.  Composition  of  any  Number  of  Forces  in  One 
Plane,  all  applied  at  the  Same  Point. 

(a)  GRAPHICAL  SOLUTION.  —  Let  the  forces  be  represented 
{Fig.  2)  by  AB,  AC,  AD,  AE,  and  AF  respectively.     Draw  Be 
||  and  —  AC,  cd  \\  and  =  AD,  de  ||  and  =  AE,  and  ef  ||  and  = 
AF;   then  will  Af  represent  the  resultant  of  the  five  forces. 
This   solution   is  to  be  deduced   from 

§  17  in  the  same  way  as  §  19  is  deduced 
from  §  1  6.  c, 

(b)  ALGEBRAIC     SOLUTION.  —  Let 
the    given    forces    (Fig.    9),    of    which    B, 
three  are  represented  in  the  figure,  be 

Ft  Fs,  F2,  Fz,  F4,  etc.  ;  and  let  the  angles 

made  by  these  forces  with  the  axis  OX    o1^  i      B    A 

be   a,   a,,   a2,  a3,  a4,    etc.,   respectively.  FlG-9- 

Resolve   each    of   these   forces   into   two   components,  in  the 

directions  OX  and  OY  respectively.     We  shall  obtain  for  the 

components  along  OX 

OA  =  Fcosa,     OB   —  Fscosat,     OC  —  F2cosa2,     etc.; 
and  for  those  along  O  Y 

OAI  =  ^sina,     OB  i  =  ^sina^     OCt  =  F2  sin  03,     etc. 


These  forces  are  equivalent  to  the  following  two  ;  viz.,  a 
force  Fcos  a  +  Fl  cos  a,  +  F2  cos  a2  +  ^3  cos  «3  +  etc-  along  OX, 
and  a  force  Fsin  a  +  Fl  sin  ax  +  F2  sin  a2  +  -^3  sin  a3  +  etc.  along 
O  Y.  The  first  may  be  represented  by  ^Fcos  a,  and  the  second 
by  'SFsma,  where  S  stands  for  algebraic  sum.  There  remains 
only  to  find  the  resultant  of  these  two,  the  magnitude  of  which 
is  given  by  the  equation 

R  = 


22 


APPLIED   MECHANICS. 


and,  if  we  denote  by  ar  the  angle  made  by  the  resultant  with 
OX,  we  shall  have 

.             ^F  sin  a 
,     smcv  =  . 

R  R 


EXAMPLES. 


i°.  Given 


a  =  21' 
a,=  48° 
a2  =  82° 


Find  the  result- 
ant force  and 
its  direction. 


Solution. 


F. 

a. 

COS  a. 

sin  a. 

F  COS  a. 

Fs'ma. 

47 

21° 

0-93358 

o.35837 

43.87826 

16.84339 

73 

48° 

0.66913 

o-743i5 

48.84649 

54-24995 

43 

82° 

O.I39I7 

0.99027 

5-98431 

42.58161 

23 

112° 

-0.37461 

0.92718 

-8.6l603 

21.32414 

90.09303 

134.99909 

a  =  90.09303, 


a  =  134.99909, 


sin  a)2  =  162.2976. 


log  ^F  COS  a    =    1.954691 

logj?  =  2.210331 


log  COS  Or  =   9.744360 

a,  =    56°  I7'. 

OBSERVATION.  —  It  would  be  perfectly  correct  to  use  the  minus  sign 
in  extracting  the  square  root,  or  to  call  R  =  —162.2976  ;  but  then  we 
should  have 


COS  Or  — 


or 


—  162.2976 
ar=   180°+  56° I i 


and 


sin «    _     134.99909 
sin  a?  —  —  — — , 

—  162.2976 


=  236 


COMPOSITION  OF  FORCES  APPLIED   AT  SAME   POINT.      2$ 

a  result  which,  if  plotted,  would  give  the  same  force  as  when  we  call 
R  =  162.2976     and     ar  =  56°  17'. 

Hence,  since  it  is  immaterial  whether  we  use  the  plus  or  the  minus  sign 
in  extracting  the  square  root  provided  the  rest  of  the  computation  be 
consistent  with  it,  we  shall,  for  convenience,  use  always  plus. 

2°.  F  =    4,  «  =     77°, 

^i=     3,  «.=     82"> 

F2=   10,  a2=   163°, 

<F3=    5,  «s=  275°. 

3°.  F  =    5,  a  =  cos -^, 

-Fi  =4,  «i  =  o, 

F2  =3,  a2  =  90°. 

§  25.  Polygon  of  Forces.  —  If  any  number  of  forces  be 
represented  in  magnitude  and  direction  by  the  sides  of  a  polygon 
taken  in  order,  then,  if  these  forces  be  simultaneously  applied  at 
one  point,  they  will  balance  each  other. 

Conversely,  any  number  of  forces  which,  when  simultaneously 
applied  at  one  point,  balance  each  other,  can  be  correctly  repre- 
sented in  magnitude  and  direction  by  the  sides  of  a  polygon  taken 
in  order. 

These  propositions  are  to  be  deduced  from  §  24  (a)  in  the 
same  way  as  the  triangle  of  forces  is  deduced  from  the  parallelo- 
gram of  forces. 

§  26.  Composition  of  Forces  all  applied  at  the  Same 
Point,  and  not  confined  to  One  Plane. — This  problem  can 
be  solved  by  the  polygon  of  forces,  since  there  is  nothing  in 
the  demonstration  of  that  proposition  that  limits  us  to  a  plane 
rather  than  to  a  gauche  polygon. 

The  following  method,  however,  enables  us  to  determine 
algebraic  values  for  the  magnitude  of  the  resultant  and  for  its 
direction. 


APPLIED   MECHANICS. 


FIG.  10. 


We  first  assume  a  system  of  three  rectangular  axes,  OX, 

OY,  and  OZ  (Fig.  10),  whose  origin 
is  at  the  common  point  of  the  given 
forces.  Now,  let  OE  —  F  be  one 
of  the  given  forces.  First  resolve 
it  into  two  forces,  OC  and  OD,  the 
first  of  which  lies  in  the  z  axis,  and 
the  second  perpendicular  to  OZ, 
or,  as  it  is  usually  called,  in  the  z 
plane ;  the  plane  perpendicular  to 
OX  being  the  x  plane,  and  that 
perpendicular  to  O  Y  the  y  plane. 
Then  resolve  OD  into  two  com- 
ponents, OA  along  OX,  and  OB  along  OY.  ,We  thus  obtain 
three  forces,  OA,  OB,  and  OC  respectively,  which  are  equivalent 
to  the  single  force  OE.  These  three  components  are  the  edges 
of  a  rectangular  parallelepiped,  of  which  OE  =  Fis  the  diagonal. 
Let,  now, 

'  angle  BOX  =  a,     EOY  =  ft     and     EOZ  =  y ; 

and  we  have,  from  the  right-angled  triangles  EOA,  EOB,  and 
EOC  respectively, 

OA  =  Fcosa,     O#  =  .Fcosft     OC  =  Fcosy. 
Moreover, 

OA*  -f  OB*  =  OD2  and  OD2     +  OC2    =  OE2 
.'.     OA2     +  OB2     +  OC2    =  OE2, 

and  by  substituting  the  values  of  OA,  OB,  and  OC,  given  above, 

we  obtain 

cos2  a  +  cos2  /3  -f  cos2 y  =  i  ; 

a  purely  geometrical  relation  existing  between  the  three  angles 
that  any  line  makes  with  three  rectangular  co-ordinate  axes. 

When  two  of  the  angles  a,  ft  and  y  are  given,  the  third  can 
be  determined  from  the  above  equation. 


COMPOSITION  OF  FORCES  APPLIED   AT  SAME   POINT.      2$ 


Resolve,  in  the  same  way,  each  of  the   given  forces  into 
three  components,  along  OX,  OY,  and  OZ  respectively,  and  we 
shall  thus  reduce  our  entire  system 
of  forces   to   the   following  three 
forces  :  — 

i°.  A  single  force  XFcosa  along  OX. 
2°.  A  single  force  XFcos/3  along  OY. 
3°.  A  single  force  2/^cosy  along  OZ. 

We  next  proceed  to  find  a  sin- 
gle resultant  for  these  three  forces. 
Let  (Fig.  ii) 


FIG.  ii. 


OA  = 

OB  =  ^F  cos  ft 

OC  =  XFcosy. 


Compounding  OA  and  OB,  we  find  OD  to  be  their  resultant ; 
and  this,  compounded  with  OC,  gives  OE  as  the  resultant  of 
the  entire  system.  Moreover, 

OE2  =  OD2  4-  OC2  =  OA2  -f-  OB2  +  OC2, 
or 

R2      =  (ZFcosa)2  -f  (XFcosft)2  4- 


and  if  we  let 
have 


R       =  V(2^cosa)2  +  (2^  cos/2)2  + 
=  ar,  EOY  =  pr,  and 


cosy)2; 
=  yr}  we  shall 


OA        ^F  cos  a  XFcos/3 

cos  ar  =  —  —  =  -  -  —  ,   cos/3r  =  -  ^—  -,   and  cosyr 

K  K  K 


This  gives  us  the  magnitude  and  direction  of  the  resultant. 

The  same  observation  applies  to  the  sign  of  the  radical  for 
R  as  in  the  case  of  forces  confined  to  one  plane. 


26  APPLIED   MECHANICS. 

DETERMINATION   OF   THE   THIRD   ANGLE   FOR   ANY   ONE   FORCE. 

When  two  of  the  angles  a,  ft,  and  y  are  given,  the  cosine  of 
the  third  may  be  determined  from  the  equation,  — 

cos2  a  +  cos2/?  +  cos2  y  =  i  ; 

but,  as  we  may  use  either  the  plus  or  the  minus  sign  in  extract- 
ing the  square  root,  we  have  no  means  of  knowing  which  of 
the  two  supplementary  angles  whose  cosine  has  been  deduced 
is  to  be  used. 

Thus,  suppose  a  =  45°,  j3  =  60°,  then 


cosy  =  ±V/i  -  -i  -  J  =  ±£ 
/.    y  =  60°,  or  1 20° ; 

but  which  of  the  two  to  use  we  have  no  means  of  deciding. 

This  indetermination  will  be  more  clearly  seen  from  the  fol- 
lowing geometrical  considerations  :  — 

The  angle  a  (Fig.  12),  being  given   as  45°,  locates  the  line 

representing  the  force  on  a  right 
circular  cone,  whose  axis  is  OXy 
and  whose  semi-vertical  angle  is 
AOX=BOX=4$°.  On  the  other 
hand,  the  statement  that  ft  =  60° 
locates  the  force  on  another  right 
circular  cone,  having  O  Y  for  axis, 
and  a  semi-vertical  angle  of  60° ; 
both  cones,  of  course,  having  their 
vertices  at  O.  Hence,  when  a  and 

FIG.  12. 

ft  are  given,  we  know  that  the  line 

representing  the  force  is  an  element  of  both  cones  ;  and  this  is 
all  that  is  given. 

(a)  Now,  if  the  sum  of  the  two  given  angles  is  less  than 
90°,  the  cones  will  not  intersect,  and  the  data  are  consequently 
inconsistent. 


DETERMINATION  OF   THE    THIRD   ANGLE.  2/ 

(b)  If,  on  the  other  hand,  one  of  the  given  angles  being 
greater  than  90°,  their  difference  is  greater  than  90°,  the  cones 
will  not  intersect,  and  the  data  are  again  inconsistent. 

(c)  If  a  +  ft  =  90°,  the  cones  are  tangent  to  each  other, 
and  y  =  90°. 

(d)  If  a  -f  /?  >  90°,  and  a  —  ft  or  ft  —  a  <  90°,  the  cones 
intersect,  and  have  two  elements  in  common ;  and  we  have  no 
means  of  determining,  without  more  data,  which  intersection 
is  intended,  this  being  the  indetermination  that  arises  in  the 
algebraic  solution. 


i.  Given 


=     2 


EXAMPLES. 


a  =  53 
a  =  87° 


=  72 


70°    y  =  45C 


Find  the  magnitude 
and  direction  of 
the  resultant. 


Solution. 


F 

a. 

ft 

Y- 

COS  a. 

COS0. 

COSy. 

F  COS  a. 

F  cos  ft. 

F  COS  y. 

63 
49 

2 

53° 
87° 

42° 
70° 

72° 

45° 

O.6OI82 
0.05234 
0.6l888 

0.74314 
0.94961 
0.34202 

0.29250 
0.30902 
0.70711 

37.91466 

2.56466 
1.23776 

46.81782 
46.53089 
0.68404 

18.42750 
15.14198 

1.41422 

41.71708 

2/^cos  a 

94.03275 
2^cos  li 

34.98370 

2F  cos  y 

R  =  V(S^cosa)2  +  C%J?  cos  ft)2  +  (X^cosy)2  =  108.6569. 
log  2/^cosa  =  1.620314       log  SjFcos/3  =  1-973279       log  2/^cosy  =  1.543866 
log  R  —  2.036057       log-/?  =  2.036057       log./?  =  2.036057 

log  cos  ar      =9.584257       log  cos /3r      =9.937222      log  cos  yr      =9.507809 
ar  =  67°  25'  20"  J3r  '  =3o°4/i4"yr  =7i°i3/5" 


28 


APPLIED   MECHANICS. 


P. 

a. 

& 

/?. 

a. 

ft 

y- 

2. 

4-3 

47°  2' 

65°  7' 

3- 

5 

90° 

90° 

87.5 

88°  3' 

10°  5' 

7 

0° 

6.4 

68°  4' 

83°    2' 

4 

0° 

75 

73° 

45° 

§  27.  Conditions  of  Equilibrium  for  Forces  applied  at  a 
Single  Point. 

i°.  When  the  forces  are  not  confined  to  one  plane,  we  have 
already  found,  for  the  square  of  the  resultant, 


2  +  (S^cos/?)2  +  (S^cosy)2. 
But  this  expression  can  reduce  to  zero  only  when  we  have 
a  =  o,     X/^cos/?  =  o,     and     S^cosy  =  o; 


for  the  three  terms,  being  squares,  are  all  positive  quantities, 
and  hence  their  sum  can  reduce  to  zero  only  when  they  are 
separately  equal  to  zero. 

Hence  :  If  a  set  of  balanced  forces  applied  at  a  single  point 
be  resolved  into  components  along  three  directions  at  right  angles 
to  each  other,  the  algebraic  sum  of  the  components  of  the  forces 
along  each  of  the  three  directions  must  be  equal  to  zero,  and  con- 
versely. 

2°.  When  the  forces  are  all  confined  to  one  plane,  let  that 
plane  be  the  z  plane  ;  then  y  =  90°  in  each  case,  and 

.-.    /?        =  9o°  -  a 
/.     cos     —  sin  a 


-h 
Hence,  for  equilibrium  we  must  have 

+  (S^sin-a)2  =  o; 


STATICS  OF  RIGID  BODIES.  29 

and,  since  this  is  the  sum  of  two  squares, 

a  =  o,  and  2LFsina  =  o. 


Hence  :  If  a  set  of  balanced  forces,  all  situated  in  one  plane, 
and  acting  at  one  point,  be  resolved  into  components  along  tivo 
directions  at  right  angles  to  each  other,  and  in  their  own  plane, 
the  algebraic  sum  of  the  components  along  each  of  the  tivo  given 
directions  must  be  equal  to  zero  respectively;  and  conversely. 

§  28.  Statics  of  Rigid  Bodies.  —  A  rigid  body  is  one  that 
does  not  undergo  any  alteration  of  shape  when  subjected  to 
the  action  of  external  forces.  Strictly  speaking,  no  body  is 
absolutely  rigid  ;  but  different  bodies  possess  a  greater  or  less 
degree  of  rigidity  according  to  the  material  of  which  they  are 
composed,  and  to  other  circumstances.  When  a  force  is  ap- 
plied to  a  rigid  body,  we  may  have  as  the  result,  not  merely  a 
rectilinear  motion  in  the  direction  of  the  force,  but,  as  will  be 
shown  later,  this  may  be  combined  with  a  rotary  motion  ;  in 
short,  the  criterion  by  which  we  determine  the  ensuing  motion 
is,  that  the  effect  of  the  force  will  distribute  itself  through  the 
body  in  such  a  way  as  not  to  interfere  with  its  rigidity. 

What  this  mode'  of  distribution  is,  we  shall  discuss  here- 
after ;  but  we  shall  first  proceed  to  some  propositions  which  can 
be  proved  independently  of  this  consideration. 

§  29.  Principle  of  Rectilinear  Transferrence  of  Force  in 
Rigid  Bodies.  —  If  a  force  be  applied  to  a  rigid  body  at  the 
point  A  (Fig.  13)  in  the  direction  AB, 
whatever  be  the  motion  that  this  force 
would  produce,  it  will  be  prevented  from 
taking  place  if  an  equal  and  opposite 
force  be  applied  at  A,  B,  C,  or  D,  or  at 
any  point  along  the  line  of  action  of  the  force  :  hence  we  have 
the  principle  that  — 

The  point  of  application  of  a  force  acting  on  a  rigid  body, 
may  be  transferred  to  any  other  point  which  lies  in  the  line  of 


UHI7ERSITT 


APPLIED  MECHANICS. 


action  of  the  force,  and  also  in  the  body,  ^vitho^lt  altering  the 
resulting  motion  of  the  body,  although  it  does  alter  its  state  of 
stress. 

§  30.  Composition  of  two  Forces  in  a  Plane  acting  at 
Different  Points  of  a  Rigid  Body,  and  not  Parallel  to  Each 
Other.  — Suppose  the  force  F  (Fig.  14)  to  be  applied  at  A,  and 
Fj  at  B,  both  in  the  plane  of  the  paper,  and  acting  on  the  rigid 
body  abcdef.  Produce  the  lines  of  direction  of  the  forces  till 
they  meet  at  O,  and  suppose  both  F  and  Fl  to  act  at  O.  Con- 
struct the  parallelogram  ODHE,  where  OD  —  F  and  OE  =  F, ; 

then  will  OH  —  R  rep- 
resent the  resultant 
force  in  magnitude  and 
in  direction.  Its  point 
of  application  may  be 
conceived  at  any  point 
along  the  line  OH,  as 
at  C,  or  any  other 
point  ;  and  a  force 
equal  and  opposite  to 
OH,  applied  at  any  point  of  the  line  OH,  will  balance  F  at  A, 
and  F.  at  B. 

The  above  reasoning  has  assumed  the  points  A,  B,  C  and 
(9,  all  within  the  body  :  but,  since  we  have  shown,  that  when 
this  is  the  case,  a  force  equal  and  opposite  to  R  at  C  will  bal- 
ance Fat  A,  and  Fl  at  B,  it  follows,  that  were  these  three  forces 
applied,  equilibrium  would  still  subsist  if  we  were  to  remove 
the  part  bafegho  of  the  rigid  body ;  or,  in  other  words,  — 

The  same  construction  holds  even  when  the  point  O  falls  out- 
side the  rigid  body. 

§31.  Moment  of  a  Force  with  Respect  to  an  Axis  Per- 
pendicular to  the  Force. 

DEFINITION.  —  The  moment  of  a  force  with  respect  to  an 
axis  perpendicular  to  the  force,  and  not  intersecting  it,  is  the 


FIG.  14. 


EQUILIBRIUM  OF   THREE  PARALLEL   FORCES. 


FIG.  15. 


product  of  the  force  by  the  common  perpendicular  to  (shortest 
distance  between)  the  force  and  the  axis. 

Thus,  in  Fig.  15  the  moment  of  F  about 
an  axis  through  O  and  perpendicular  to  the 
plane  of  the  paper  is  F(OA).  The  sign  of 
the  moment  will  depend  on  the  sign  attached 
to  the  force  and  that  attached  to  the  perpen- 
dicular. These  will  be  assumed  in  this  book 
in  such  a  manner  as  to  render  the  following  true  ;  viz,,  — 

The  moment  of  a  force  with  respect  to  an  axis  is  called  posi- 
tive when,  if  the  axis  ivere  supposed  fixed,  the  force  woztld  cause 
the  body  on  which  it  acts  to  rotate  around  the  axis  in  the  direc- 
tion of  the  hands  of  a  watch  as 
seen  by  the  observer  looking  at 
the  face.  It  will  be  called  nega- 
tive zvhen  the  rotation  would  take 
place  in  the  opposite  direction. 

§  32.  Equilibrium  of  Three 
Parallel  Forces  applied  at 
Different  Points  of  a  Rigid 
Body.  —  Let  it  be  required  to 
find  a  force  (Fig.  16)  that  will 
balance  the  two  forces  F  at  A, 
and  Ft  at  B.  Apply  at  A  and  B 
respectively,  and  in  the  line  AB, 
the  equal  and  opposite  forces  Aa 
and  Bb.  Their  introduction  will 
produce  no  alteration  in  the 
body's  motion. 

The  resultant  of  F  and  Aa 
is  Af,  that  of  F,  and  Bb  is  Bg. 
Compound  these  by  the  method 
of  §  30,  and  we  obtain  as  result- 
ant ce.     A  force  equal  in  magnitude  and  opposite  in  direction 


FIG.  16. 


APPLIED   MECHANICS. 


to  ce,  applied  at  any  point  of  the  line  cC,  will  be  the  force 
required  to  balance  Fat  A  and  Fl  at  B ;  and,  as  is  evident  from 
the  construction,  this  line  is  in  the  plane  of  the  two  forces. 
Moreover,  by  drawing  triangle  fKl  equal  to  Bbg,  we  can  readily 
prove  that  triangles  oce  and  Afl  are  equal :  hence  the  angle  oce 
equals  the  angle  fAl,  and  R  is  parallel  to  Fand  Ft.  Also 

R  =  ce  =  ch  4-  he  =  AK  -f  Kl  =  F  -f  Flt 


and 


Cc  =  AK  =  JF 

^c     /&:  ~  ^0' 
CL~  M.-1L. 

BC       Bb       Bb* 


7i*  7i*  7?  7?      i         T?  J?  E1  77* 

^C  ""  y  "  ~BC  =  ZC  = 


snce     #  = 


AB 


Cr        Cq 


qr 


where  qr  is  any  line  passing  through  C. 

Hence  we  have  the  following  propositions  ;  viz.,  — 
If  three  parallel  forces  balance  each  other,  — 
1°.    They  must  lie  in  one  plane. 

2°.    The  middle  one  must  be  equal  in  magnitude  and  opposite 
FI  in  direction  to  the   sum   of  the  other 

two. 

3°.  Each  force  is  proportional  to  the 
o    distance  between  the  lines  of  direction 
of  the  other  two  as  measured  on  any 
line  intersecting  all  of  them. 

The  third  of  the  above-stated  con- 
ditions may  be  otherwise  expressed, 
thus : — 

FIG.  17.  The  algebraic  sum  of  the  moments 

of  the  three  forces  about  any  axis  perpendicular  to  the  forces 
must  be  zero. 


RESULTANT  OF  A    PAIR   OF  PARALLEL   FORCES.  33 

PROOF.  —  Let  F,  Flt  and  R  (Fig.  17)  be  the  forces  ;  and  let 
the  axis  referred  to  pass  through  O.  Draw  OA  perpendicular 
to  the  forces.  Then  we  have 


F(OA)  +  Fi(OB)  =  F(OC  +  CA)  +  FS(OC  -  BC) 
=  (F  +  Ft)OC     +  F(AC)  - 

But,  from  what  we  have  already  seen, 

F  +  Ft  =  -R 

and 

JL       J± 
BC      AC 

,%     ^(^C)  =  FS(BC) 

.-.    F(OA)  +Ft(OB)  =  -R(OC)  +o 

.-.     F(OA)  +  Ft  (OB)  +  ^(0C)  =  o, 


or  the  algebraic  sum  of  the  moments  of  the  forces  about  the 
axis  through  O  is  equal  to  zero. 

§  33.  Resultant  of  a  Pair  of  Parallel  Forces.  —  In  the 
preceding  case,  the  resultant  of  any  two  of  the  three  forces 
Fy  Flt  and  R,  in  Fig.  16  or  Fig.  17,  is  equal  and  opposite  to  the 
third  force.  Hence  follow  the  two  propositions  :  — 

I.  If  two  parallel  forces  act   in   the   same  direction,  their 
resultant  lies  in  the  plane  of  the  forces,  is  equal  to  their  sum, 
acts  in  the  same  direction,  and  cuts  the  line  joining  their  points 
of  application,  or  any  common  perpendicular  to  the  two  forces, 
at  a  point  which  divides  it  internally  into  two  segments  in- 
versely as  the  forces. 

II.  If  two  unequal  parallel  forces  act  in  opposite  directions, 
their  resultant  lies  in  the  plane  of  the  forces,  is  equal  to  their 
difference,  acts  in  the  direction  of  the  larger  force,  and  cuts  the 
line  joining  their  points  of  application,  or  any  common  perpen- 
dicular to  them,  at  a  point  which  (lying  nearer  the  larger  force) 


34  APPLIED   MECHANICS. 

divides  it  externally  into  two  segments  which  are  inversely  as 
the  forces. 

Another  mode  of  stating  the  above  is  as  follows  :  — 

i°.  The  resultant  of  a  pair  of  parallel  forces  lies  in  the  plane 
of  the  forces. 

2°.  It  is  equal  in  magnitude  to  their  algebraic  sum,  and  coin- 
cides in  direction  with  the  larger  force. 

3°.  The  moment  of  the  resultant  about  an  axis  perpendicu- 
lar to  the  plane  of  the  forces  is  equal  to  the  algebraic  sum  of 
the  moments  about  the  same  axis. 

EXAMPLES. 

1.  Find  the  length  of  each  arm  of  a  balance  such  that  i  ounce  at 
the  end  of  the  long  arm  shall  balance  i  pound  at  the  end  of  the  short 
arm,  the  length  of  beam  being  2  feet,  and  the  balance  being  so  propor- 
tioned as  to  hang  horizontally  when  unloaded. 

2.  Given  beam  =  28  inches,  3  ounces  to  balance  15. 

3.  Given  beam  =  36  inches,  5  ounces  to  balance  25  ounces. 

MODE  OF  DETERMINING  THE  RESULTANT  OF  A  PAIR  OF  PARALLEL 
FORCES  REFERRED  TO  A  SYSTEM  OF  THREE -RECTANGULAR 
AXES. 

Let  both  forces  (Fig.  18)  be  parallel  to  OZ ;  tijjpi  we  have, 
from  what  has  preceded, 

F_Fl_F-\-Fl_R      ,        R  —  F  -\-  F 
be       ab  ac  ac 

But  from  the  figure 

be  ab          .          F       _       F, 

X$  ~~  X2          X2          Xi 

.'.     Fx2  —  FxL   = 


or 

Rx2  =  Fxt 


RESULTANT  OF  NUMBER   OF  PARALLEL   FORCES.          35 


and  similarly  we  may  prove  that 
Ry2  =  Fy* 


or 


i°.  The  resultant  of  two  parallel  forces  is  parallel  to  the 
forces  and  equal  to  their  algebraic  sum. 


R=F+Ft 


FIG.  18. 

2°.  The  moment  of  the  resultant  with  respect  to  OX  is 
equal  to  the  algebraic  sum  of  their  moments  with  respect  to 
OX ' ;  and  likewise  when  the  moments  are  taken  with  respect 
to  OY. 

§  34.  Resultant  of  any  Number  of  Parallel  Forces.  — 
Let  it  be  required  to  find  the  resultant  of  any  number  of  paral- 
lel forces. 

In  any  such  case,  we  might  begin  by  compounding  two  of 
them,  and  then  compounding  the  resultant  of  these  two  with  a 
third,  this  new  resultant  with  a  fourth,  and  so  on.  Hence,  for 
the  magnitude  of  any  one  of  these  resultants,  we  simply  add 
to  the  preceding  resultant  another  one  of  the  forces ;  and  for 
the  moment  about  any  axis  perpendicular  to  the  forces,  we  add 


APPLIED   MECHANICS 


to  the  moment  of  the  preceding  resultant  the  moment  of  the 
new  force. 

Hence  we  have  the  following  facts  in  regard  to  the  resultant 
of  the  entire  system  :  — 

I  °.  The  resultant  will  be  parallel  to  the  forces  and  equal  to 
their  algebraic  sum. 

2°.  The  moment  of  the  resultant  about  any  axis  perpendicular 
to  the  forces  will  be  eqtial  to  the  algebraic  sum  of  the  moments 
of  the  forces  about  the  same  axis. 

The  above  principles  enable  us  to  determine  the  resultant 
in  all  cases,  except  when  the  algebraic  sum  of  the  forces  is 
equal  to  zero.  This  case  will  be  considered  later. 

§  35.   Composition    of    any    System    of    Parallel    Forces 

when  all  are  in  One  Plane. — 
Refer  the  forces  to  a  pair  of  rect- 
angular axes,  OX,  OY  (Fig.  19), 
and  assume  OY  parallel  to  the 
forces. 

The  forces  and  the  co-ordinates 
of  their  lines  of  direction  being  as 
indicated  in  the  figure,  if  we  denote 
0  by  R  the  resultant,  and  by  XQ  the 
co-ordinate  of  its  line  of  direction, 
we  shall  have,  from  the  preceding, 

R  =  ^F;  (i) 

and  if  moments  be  taken  about  an 
axis  through  O,  and  perpendicular 


F,     F, 


FIG.  19. 


to  the  plane  of  the  forces,  we  shall  also  have 


Hence 


(2) 


R  = 


and     xn  = 


determine  the  resultant  in   magnitude  and  in  line  of   action, 
except  when  %F  =  o,  which  case  will  be  considered  later. 


EQUILIBRIUM  OF  ANY  SET  OF  PARALLEL   FORCES.       37 

§  36,  Composition  of  any  System  of  Parallel  Forces  not 
confined  to  One  Plane.  —  Refer  the  forces  to  a  set  of  rect- 
angular axes  so  chosen  that  OZ  is  parallel  to  their  direction. 
If  we  denote  the  forces  by  Flt  F21  F3,  F4,  etc.,  and  the  co-ordinates 
of  their  lines  of  direction  by  (#„  jj/,),  (x^  y2),  etc.,  and  if  we 
denote  their  resultant  by  R,  and  the  co-ordinates  of  its  line  of 
direction  by  (xm  y0),  we  shall  have,  in  accordance  with  what  has 
been  proved  in  §  34,  — 

i°.  The  magnitude  of  the  resultant  is  equal  to  the  algebraic 
sum  of  the  forces,  or 

R  =  ^F. 

2°.  The  moment  of  tJie  resultant  about  OY  is  equal  to  the 
sum  of  the  moments  of  the  forces  about  OY,  or 


3°.    The  moment  of  the  resultant  about  OX  is  equal  to  the 
sum  of  the  moments  about  OX,  or 


Hence 


determine  the  resultant  in  all  cases,  except  when  ^F  =  o. 

§  37.  Conditions  of  Equilibrium  of  any  Set  of  Parallel 
Forces.  —  If  the  axes  be  assumed  as  before,  so  that  OZ  is 
parallel  to  the  forces,  we  must  have 


=  o,     2Fx  =  o,     and     2Fy  =  o. 


To  prove  this,  compound  all  but  one  of  the  forces.  Then  equilib- 
rium will  subsist  only  when  the  resultant  thus  obtained  is  equal 
and  directly  opposed  to  the  remaining  force  ;  i.e,  it  must  be 
equal,  and  act  along  the  same  line  and  in  the  opposite  direction. 
Hence,  calling  Ra  the  resultant  above  referred  to,  and  (xa,  ya) 
the  co-ordinates  of  its  line  of  direction,  and  calling  FH  the 


38  APPLIED   MECHANICS. 

remaining  force,  and  (xn,  y^)  the  co-ordinates  of  its  line  of  direc- 
tion, we  must  have 

Ra  =    —  FH,      Xa  =   Xnt       ya  =  )'n, 

.'.    Ra  +  Fn  =      o,       Raxa  +  Fnxn  —  o,       Raya  -f  Fuyn  =  o, 
.-.    2F  =      o,       *ZFx  =  o,       2/3>  =  o. 


When  the  forces  are  all  in  one  plane,  the  conditions  become 
2F  =  o,     ^Fx  =  o. 

§  38.   Centre    of    a    System    of   Parallel    Forces.  —  The 

resultant  of  the  two  parallel  forces  F  and  Ft  (Fig.  20),  ap- 
plied at  A  and  B  respectively,  is  a  force  R  =  F  +  Flt  whose 
line  of  action  cuts  the.  line  AB  at  a  point  C, 
which  divides  it  into  two  segments  inversely  as 
the  forces.  If  the  forces  F  and  F,  are  turned 
through  the  same  angle,  and  assume  the  posi- 
tions AO  and  BOl  respectively,  the  line  of 
action  of  the  resultant  will  still  pass  through 
C,  which  is  called  the  centre  of  the  two  parallel 
forces  F  and  Ft.  Inasmuch  as  a  similar  reasoning  will  apply 
in  the  case  of  any  number  of  parallel  forces,  we  may  give  the 
following  definition  :  — 

The  centre  of  a  system  of  parallel  forces  is  the  point  through 
which  the  line  of  action  of  the  resultant  always  passes,  no  matter 
how  the  forces  are  turned,  provided  only  — 

1°.    Their  points  of  application  remain  the  same. 
2°.    Their  relative  magnitudes  are  unchanged. 
3°.   They  remain  parallel  to  each  other. 

Hence,  in  finding  the  centre  of  a  set  of  parallel  forces,  we 
may  suppose  the  forces  turned  through  any  angle  whatever,  and 
the  centre  of  the  set  is  the  point  through  which  the  line  of 
action  of  the  resultant  always  passes. 


DISTRIBUTED   FORCES. 


39 


§  39-  Co-ordinates  of  the  Centre  of  a  Set  of  Parallel 
Forces.  —  Let  Fl  (Fig.  21)  be  one  of  the  forces,  and  (xlt  yiy  zt) 
the  co-ordinates  of  its  point 
of  application.  Let  F2  be 
another,  and  (x2,  y2,  z2)  co- 
ordinates of  its  point  of 
application.  Turn  all  the 
forces  around  till  they  are 
parallel  to  OZ,  and  find  the 
line  of  direction  of  the  re- 
sultant force  when  they  are 
in  this  position.  The  co- 
ordinates of  this  line  are 

FIG.  21. 


and,  since  the  centre  of  the  system  is  a  point  on  this  line,  the 
above  are  two  of  the  co-ordinates  of  the  centre.  Then  turn 
the  forces  parallel  to  OX,  and  determine  the  line  of  action  of 
the  resultant.  We  shall  have  for  its  co-ordinates 


7o  = 


Hence,  for  the  co-ordinates  of  the  centre  of  the  system,  we 
have 


When  2.F  —  o  the  co-ordinates  would  be  oo,  therefore  such 
a  system  has  no  centre. 

§  40.  Distributed  Forces While  we  have  thus  far  as- 
sumed our  forces  as  acting  at  single  points,  no  force  really  acts 
at  a  single  point,  but  all  are  distributed  over  a  certain  surface 


40  APPLIED   MECHANICS. 

or  through  a  certain  volume ;  nevertheless,  the  propositions 
already  proved  are  all  applicable  to  the  resultants  of  these 
distributed  forces.  We  shall  proceed  to  discuss  distributed 
forces  only  when  all  the  elements  of  the  distributed  force  are 
parallel  to  each  other.  As  a  very  important  example  of  such  a 
distributed  force,  we  may  mention  the  force  of  gravity  which 
is  distributed  through  the  mass  of  the  body  on  which  it  acts. 
Thus,  the  weight  of  a  body  is  the  resultant  of  the  weights  of 
the  separate  parts  or  particles  of  which  it  is  composed.  As 
another  example  we  have  the  following :  if  a  straight  rod  be 
subjected  to  a  -direct  pull  in  the  direction  of  its  length,  and  if 
it  be  conceived  to  be  divided  into  two  parts  by  a  plane  cross- 
section,  the  stress  acting  at  this  section  is  distributed  over  the 
surface  of  the  section. 

§41.  Intensity  of  a  Distributed  Force.  —  Whenever  we 
have  a  force  uniformly  distributed  over  a  certain  area,  we  obtain 
its  intensity  by  dividing  its  total  amount  by  the  area  over  which 
it  acts,  thus  obtaining  the  amount  per  unit  of  area. 

If  the  force  be  not  uniformly  distributed,  or  if  the  intensity 
vary  at  different  points,  we  must  adopt  the  following  means 
for  finding  its  intensity.  Assume  a  small  area  containing  the 
point  under  consideration,  and  divide  the  total  amount  of  force 
that  acts  on  this  small  area  by  the  area,  thus  obtaining  the 
mean  intensity  over  this  small  area :  this  will  be  an  approxima- 
tion to  the  intensity  at  the  given  point;  and  the  intensity  is  the 
limit  of  the  ratio  obtained  by  making  the  division,  as  the  area 
used  becomes  smaller  and  smaller. 

Thus,  also,  the  intensity,  at  a  given  point,  of  a  force  which 
is  distributed  through  a  certain  volume,  is  the  limit  of  the 
ratio  of  the  force  acting  on  a  small  volume  containing  the 
given  point,  to  the  volume,  as  the  latter  becomes  smaller  and 
smaller. 

§42.  Resultant  of  a  Distributed  Force. —  i°.  Let  the 
force  be  distributed  over  the  straight  line  AB  (Fig.  22),  and 


RESULTANT  OF  A    DISTRIBUTED   FORCE. 


let  its  intensity  at  the  point  E  where  AE  —  x,  be  represented 
by  EF  =  /  =  <j>(x),  a  function  of  x ; 
then  will  the  force  acting  on  the  por- 
tion Ee  =  kx  of  the  line  be/A^r:  and 
if  we  denote  by  R  the  magnitude  of 
the  resultant  of  the  force  acting  on  the 
entire  line  AB,  and  by  XQ  the  distance 
of  its  point  of  application  from  A,  we  shall  have 

R  =  2/Ajc  approximately, 
or 

R  =  fpdx  exactly ; 

and,  by  taking  moments  about  an  axis  through  A  perpendicular 
to  the  plane  of  the  force,  we  shall  have 

xQR  =  2U:(/Ajt)  approximately, 
or 

xQR  =  fpxdx  exactly ; 

whence  we  have  the  equations 

fpxdx 
R  =  fpdx,  XQ  =     -pdx  ' 

2°.  Let  the  force  be  distributed  over  a  plane  area  EFGH 

(Fig.  23),  let  this  area  be  re- 
ferred to  a  pair  of  rectangular 
axes  OX  and  OY,  in  its  own 
plane,  and  let  the  intensity 
of  the  force  per  unit  of  area 
at  the  point  P,  whose  co- 
B  ordinates  are  x  and  y,  be 
p  =  <j>(x,  y)  ;  then  will  p&x&y 
be  approximately  the  force  act- 
ing on  the  small  rectangular 
area  A^-Aj.  Then,  if  we  rep- 
resent by  R  the  magnitude  of 
the  resultant  of  the  distributed  force,  and  by  x0,  yoy  the  co-ordi- 


DC 

FIG.  23. 


42  APPLIED   MECHANICS. 

nates  of  the  point  at  which  the  line  of  action  of  the  resultant 
cuts  the  plane  of  EFGH>  we  shall  have 

R  =  *Zp&xky  approximately, 
x0R  = 


or,  as  exact  equations,  we  shall  have 

R  =  ffpdxdy, 

_  ffpxdxdy  =  ffpydxdy 

~  ffpdxdy'  ~  ffpdxdy' 

3°.  Let  the  force  be  distributed  through  a  volume,  let  this 
volume  be  referred  to  a  system  of  rectangular  axes,  OX,  O  Y, 
and  OZ,  let  A  V  represent  the  elementary  volume,  whose  co- 
ordinates are  x,  y,  z,  and  let  p  =  <$>(x,  y,  2)  be  the  intensity  of 
the  force  per  unit  of  volume  at  the  point  (x,  y,  2)  ;  then,  if  we 
represent  by  R  the  magnitude  of  the  resultant,  and  by  x0,  y0,  z0, 
the  co-ordinates  of  the  centre  of  the  distributed  force,  we  shall 
have,  from  the  principles  explained  in  §  38  and  §  39,  the  approx- 
imate equations 

R  =  S/A  F,  x0R  =  2*(/A  F),  y0R  = 


and  these  give,  on  passing  to  the  limit,  the  exact  equations 


§  43.  Centre  of  Gravity.  —  The  weight  of  a  body,  or  system 
of  bodies,  is  the  resultant  of  the  weight  of  the  separate  parts 
or  particles  into  which  it  may  be  conceived  to  be  divided  ;  and 
the  centre  of  gravity  of  the  body,  or  system  of  bodies,  is  the 
centre  of  the  above-stated  system  of  parallel  forces,  i.e.,  the 
point  through  which  the  resultant  always  passes,  no  matter  how 
the  forces  are  turned.  The  weight  of  any  one  particle  is  the 
force  which  gravity  exerts  on  that  particle  :  hence,  if  we  repre- 


FORCE  APPLIED    TO   CENTRE   OF  STRAIGHT  ROD.         43 

sent  the  weight  per  unit  of  volume  of  a  body,  whether  it  be 
the  same  for  all  parts  or  not,  by  w,  we  shall  have,  as  an 
approximation, 


and  as  exact  equations, 

fwxdV  fwydV  fwzdV 

W=fwdV,     *»= 


where  W  denotes  the  entire  weight  of  the  body,  and  xot  ym  zm 
the  co-ordinates  of  its  centre  of  gravity. 

If,  on  the  other  hand,  we  let  M  =  entire  mass  of  the  body, 
dM  =  mass  of  volume  dV,  and  m  =  mass  of  unit  of  volume, 
we  shall  have 

W  —  Mg,     w  =  mg,     wdV  =mgdV  —  gdM. 
Hence  the  above  equations  reduce  to 

fxdM  fydM  fzdM 


Equations  (i)  and  (2)  are  both  suitable  for  determining  the 
centre  of  gravity;  one  of  the  sets  being  sometimes  most  con- 
venient, and  sometimes  the  other. 

§  44.  Centre  of  Gravity  of  Homogeneous  Bodies.  —  If 
the  body  whose  centre  of  gravity  we  are  seeking  is  homogeneous, 
or  of  the  same  weight  per  unit  of  volume  throughout,  we  shall 
have,  that  w  =  a  constant  in  equations  (i)  ;  and  hence  these 
reduce  to 

fxdV  fydV  fzdV 


§45.  Effect  of  a  Single  Force  applied  at  the  Centre  of 
a  Straight  Rod  of  Uniform  Section  and  Material.  —  If  a 
straight  rod  of  uniform  section  and  material  have  imparted  to  it 


44  APPLIED   MECHANICS. 

a  motion,  such  that  the  velocity  imparted  in  a  unit  of  time  to 
each  particle  of  the  rod  is  the  same,  and  if  we  represent  this 
velocity  by/,  then  if  at  each  point  of  the  rod,  we  lay  off  a  line 
xy  (Fig.  24)   in   the  direction  of   the  motion, 
and  representing  the  velocity  imparted  to  that 
point,   the  line   bounding  the   other  ends    of 
the  lines  xy  will  be  straight,  and  parallel  to  the 
rod.     If  we  conceive  the  rod  to    be  divided 
into  any  number  of   small    equal   parts,   and 


denote  the  mass  of  one  of  these  parts  by  \M,  then  will 
contain  as  many  units  of  momentum  as  there  are  units  of  force 
in  the  force  required  to  impart  to  this  particle  the  velocity 
f  in  a  unit  of  time  ;  and  hence  f^M  is  the  measure  of  this 
force. 

Hence  the  resultant  of  the  forces  which  impart  the  velocity 
f  to  every  particle  of  the  rod  will  have  for  its  measure 


where  M  is  the  entire  mass  of  the  rod  ;  and  its  point  of  applica- 
tion will  evidently  be  at  the  middle  of  the  rod. 

It  therefore  follows  that  — 

The  effect  of  a  single  force  applied  at  the  middle  of  a  straight 
rod  of  uniform  section  and  material  is  to  impart  to  the  rod  a 
motion  of  translation  in  the  direction  of  the  force,  all  points  of 
the  rod  acquiring  equal  velocities  in  equal  times. 

§  46.  Translation  and  Rotation  combined.  —  Suppose  that 
we  have  a  straight  rod  AB  (Fig.  25),  and  suppose  that  such  a 
force  or  such  forces  are  applied  to  it  as  will  impart  to  the  point 
A  in  a  unit  of  time  the  velocity  Aa,  and  to  the  point  B  the 
(different)  velocity  Bb  in  a  unit  of  time,  both  being  perpendicu- 
lar to  the  length  of  the  rod.  It  is  required  to  determine  the 
motion  of  any  other  point  of  the  rod  and  that  of  the  entire 
rod. 


TRANSLATION  AND  ROTATION  COMBIATED. 


45 


FIG.  25. 


Lay  off  Aa  and  Bb  (Fig.  25),  and  draw  the  line  ab,  and  pro- 
duce it  till  it  meets  AB  produced 
in  O:  then,  when  these  velocities 
Aa  and  Bb  are  imparted  to  the 
points  A  and  B,  the  rod  is  in  the 
act  of  rotating  around  an  axis 
through  0  perpendicular  to  the  plane  of  the  paper ;  for  when  a 
body  is  rotating  around  an  axis,  the  linear  velocity  of  any  point 
of  the  body  is  perpendicular  to  the  line  joining  the  point  in 
question  with  the  axis  (i.e.,  the  perpendicular  dropped  from  the 
point  in  question  upon  the  axis),  and  proportional  to  the  dis- 
tance of  the  point  from  the  axis. 

Hence  :  If  the  velocities  of  two  of  the  points  in  the  rod  are 
given,  and  if  these  are  perpendicular  to  the  rod,  the  motion 
of  the  rod  is  fixed,  and  consists  of  a  rotation  about  some  axis 
at  right  angles  to  the  rod. 

Another  way  of  considering  this  motion  is  as  follows  :  Sup- 
pose, as  before,  the  velocities  of  the  points  A  and  B  to  be 

represented  by  Aa  and  Bb  respec- 
tively, and  hence  the  velocity  of 
any  other  point,  as  x  (Fig.  26),  to 
be  represented  by  xy,  or  the  length 
of  the  line  drawn  perpendicular  to 
FIG.  26.  AB,  and  limited  by  AB  and  ab. 

Then,  if  we  lay  off  Aa,  —  Bb,  =  \(Aa  +  Bb}  —  Cc,  and  draw 
aj>u  and  if  we  also  lay  off  Aa2  =  ata,  and  Bb2  =  bjby  we  shall 
have  the  following  relations  ;  viz.,  — 


Aa  =  AaI  —  Aa2, 
Bb  =  Bb,  +  Bb*t 
xy  =  xy,  —  xy2, 


etc., 


or  we  may  say  that  the  actual  motion  imparted  to  the  rod  in  a 
unit  of  time  may  be  considered  to  consist  of  the  following  two 
parts :  — 


46  APPLIED   MECHANICS. 

i°.  A  velocity  of  translation  represented  by  Aaiy  the  mean 
velocity  of  the  rod ;  all  points  moving  with  this  velocity. 

2°.  A  varying  velocity,  different  for  every  different  point, 
and  such  that  its  amount  is  proportional  to  its  distance  from 
C,  the  centre  of  the  rod,  as  graphically  shown  in  the  triangles 
Aa2CBb2.  In  other  words,  the  rod  has  imparted  to  it  two 
motions  :  — 

i°.  A  translation  with  the  mean  velocity  of  the  rod. 

2°.  A  rotation  of  the  rod  about  its  centre. 

§  47.  Effect  of  a  Force  applied  to  a  Straight  Rod  of 
Uniform  Section  and  Material,  not  at  its  Centre.  —  If  the 
force  be  not  at  right  angles  to  the  rod,  resolve  it  into  two  com- 
ponents, one  acting  along  the  rod,  and  the  other  at  right  angles 
to  it.  The  first  component  evidently  produces  merely  a  trans- 
lation of  the  rod  in  the  direction  of  its  length  :  hence  the  second 
component  is  the  only  one  whose  effect  we  need  to  study. 

To  do  this  we  shall  proceed  to  show,  that,  when  such  a  rod 
has  imparted  to  it  the  motion  described  in  §  46,  the  single  re- 
A  cd  B  sultant  force  which  is  required  to  impart 

this  motion  in  a  unit  of  time  is  a  force 
acting  at  right  angles  to  the  rod,  at  a  point 
different  from  its  centre ;  and  we  shall  de- 
FIG.  27.  termine  the  relation  between  the  force  and 

the  motion  imparted,  so  that  one  may  be  deduced  from  the 
other. 

Let  A  be  the  origin  (Fig.  27),  and  let 

Ac    =  x,  cd  —  dx. 

AB  —  I  —  length  of  the  rod. 

ce  =f=  velocity  imparted  per  unit  of  time  at  distances 
from  A. 

Aa  =/„£&=/„ 

w     —  weight  per  unit  of  length. 

m     —  mass  per  unit  of   length  =  —. 


EFFECT  OF  FORCE   APPLIED    TO   A   STRAIGHT  ROD.       47 

W    =  entire  weight  of  rod. 

W 
M '  =  entire  mass  of  rod  =  — . 

g 
R     =  single  resultant  force  acting  for  a  unit  of  time  to 

produce  the  motion. 
XQ     •=.  distance  from  A  to  point  of  application  of  R. 

Then  we  shall  have, 


Hence,  from  §  42, 

(area  AabB)  =  ^(/  +  /,)/  =  ^(/  +/,).  (i) 

o  2  2 

l  m  M 

XOR  = 


,,  -_ 

" 


3   /,  +/. 

We  thus  have  a  force  -/?,  perpendicular  to  ^4^,  whose  mag- 
nitude is  given  by  equation  (i),  and  whose  point  of  application 
is  given  by  equation  (3)  ;  the  respective  velocities  imparted  by 
the  force  being  shown  graphically  in  Fig.  27. 


EXAMPLES. 

i.  Given  Weight  of  rod  =  W  =  100  Ibs., 
Length  of  rod  =  3  feet, 

Assume  £          =  32  feet  per  second, 

Force  applied  =  R  =       5  Ibs., 
Point  of  application  to  be  2.5  feet  from  one  end; 

determine  the  motion  imparted  to  the  rod  by  the  action  of  the  force  for 
one  second. 


APPLIED  MECHANICS. 


Solution. 
Equation  (i)  gives  us, 


Equation  (2)  gives, 

(2-5)  (5)  =  (^)  ©  (3)(/«  +  '/»),  or/  +  2/2  =  8 

•••    /  =  4-8,        /  =  -1.6. 

Hence  the  rod  at  the  end  nearest  the  force  acquires  a  velocity  of  4.8 
feet  per  second,  and  at  the  other  end  a  velocity  of  —1.6  feet  per 
second.  The  mean  velocity  is,  therefore,  1.6  feet  per  second;  and  we 
may  consider  the  rod  as  having  a  motion  of  translation  in  the  direc- 
tion of  the  force  with  a  velocity  of  1.6  feet  per  second,  and  a  rotation 
about  its  centre  with  such  a  speed  that  the  extreme  end  (i.e.,  a  point 
|  feet  from  the  centre)  moves  at  a  velocity  4.8  —  1.6  =  3.2  feet  per 

second.     Hence  angular  velocity  =  ^  =  2.14  per  second  =  122°. 6 

per  second. 

2.  Given  W  =  50  Ibs.,  /  =  5  feet.     It  is  desired  to  impart  to  itr 
in  one  second,  a  velocity  of  translation  at  right  angles  to  its  length,  of  5 
feet  per  second,  together  with  a  rotation  of  4  turns  per  second :  find  the 
force  required,  and  its  point  of  application. 

3.  Assume  in  example    2  that   the  velocity  of  translation    is    in   a 
direction  inclined  45°  to  the  length  of  the  rod,  instead  of  90°.     Solve 
the  problem. 

4.  Given  a  force  of  3  Ibs.  acting  for  one-half  a  second  at  a  distance 
of  4  feet  from  one  end  of  the  rod,  and  inclined  at  30°  to  the  rod  : 
determine  its  motion. 

5.  Given  the  same  conditions  as  in  example  4,  and  also   a   force 
of  4  Ibs.,  parallel  and  opposite  in  direction  to  the  3-lb.  force,  and  acting 
also  for  one-half  a  second,  and  applied  at  3  feet  from  the  other  end : 
determine  the  resulting  motion. 


MOMENT  OF   THE  FORCES   CAUSING   ROTATION.  49 

6.  Given  two  equal  and  opposite  parallel  forces,  each  acting  at  right 
angles  to  the  length  of  the  rod,  and  each  equal  to  4  Ibs.,  one  being 
applied  at  i  foot  from  one  end,  and  the  other  at  the  middle  of  the  rod ; 
find  the  motion  imparted  to  the  rod  through  the  joint  action  of  these 
forces  for  one-third  of  a  second. 

§  48.  Moment  of  the  Forces  causing  Rotation.  —  Re- 
ferring again  to  Fig.  26,  and  considering  the  motion  of  the 
rod  as  a  combination  of  translation  and  rotation,  if  we  take 
moments  about  the  centre  C,  and  compare  the  total  moment 
of  the  forces  causing  the  rotation  alone,  whose  accelerations 
are  represented  by  the  triangles  aa^cbj),  with  the  total  moment 
of  the  actual  forces  acting,  whose  accelerations  are  represented 
by  the  trapezoid  AabB,  we  shall  find  these  moments  equal  to 
each  other ;  for,  as  far  as  the  forces  represented  by  the  rectangle 
are  concerned,  every  elementary  force  m(xy^)dx  on  one  side  of 
the  centre  C  has  its  moment  (Cx]\m(xy^)dx\  equal  and  opposite 
to  that  of  the  elementary  force  at  the  same  distance  on  the  other 
side  of  C:  hence  the  total  moment  of  the  forces  represented 
graphically  by  the  rectangle  AaJj^B  is  zero,  and  hence  — 

The  moment  about  C  of  those  represented  by  the  trapezoid 
equals  the  moment  of  those  represented  by  the  triangles. 

Hence,  from  the  preceding,  and  from  what  has  been  pre- 
viously proved,  we  may  draw  the  following  conclusions  :— ; 

i°.  If  a  force  be  applied  at  the  centre  of  the  rod,  it  will 
impart  the  same  velocity  to  each  particle. 

2°.  If  a  force  be  applied  at  a  point  different  from  the  centre, 
and  act  at  right  angles  to  its  length,  it  will  cause  a  translation 
of  the  rod,  together  with  a  rotation  about  the  centre  of  the  rod. 

3°.  In  this  latter  case,  the  moment  of  the  forces  imparting 
the  rotation  alone  is  equal  to  the  moment  of  the  single  resultant 
force  about  the  centre  of  the  rod,  and  the  velocity  of  translation 
imparted  in  a  unit  of  time  is  equal  to  the  number  of  units  of 
force  in  the  force,  divided  by  the  entire  mass  of  the  rod. 


APPLIED   MECHANICS. 


§49-  Effect  of  a  Pair  of  Equal  and  Opposite  Parallel 
Forces  applied  to  a  Straight  Rod  of  Uniform  Section  and 
Material.  —  Suppose  the  rod  to  be  AB  (Fig.  28),  and  let  the 
two  equal  and  opposite  parallel  forces  be  Dd  and  Ee,  each  equal 
to  F,  applied  at  D  and  E  respectively. 
The  mean  velocity  imparted  in  a  unit 

of   time  by  either  force  will  be  —  ;  and, 


from  what  we  have  already  seen,  the  trap- 
ezoid  AabB  will  furnish  us  the  means  of 
representing  the  actual  velocity  imparted 
to  any  point  of  the  rod  by  the  force  Dd. 
The  relative  magnitudes  of  Aa  and  Bb,  the 
accelerations  at  the  ends,  will  depend,  of 
course,  on  the  position  of  D ;  but  we  shall 

always    have    Cc  —  \(Aa  -f-  Bb)  =  — ,    a 


FIG.  28. 


quantity  depending  only  on  the  magnitude 
of  the  force.  So,  likewise,  the  trapezoid  Aa.b.B  will  represent 
the  velocities  imparted  by  the  force  Ee ;  and  while  the  relative 
magnitude  of  Aa,  and  Bbl  will  depend  upon  the  position  of  E, 

c* 

we  shall  always  have  Cc,  =  \(Aa,  +  Bb,)  =  — .     Hence,  since 

Cc  =^  Cclt  the  centre  C  of  the  rod  has  no  motion  imparted  to  it 
by  the  given  pair  of  forces,  hence  the  motion  of  the  rod  is  one 
of  rotation  about  its  centre  C. 

The  resulting  velocity  of  any  point  of  the  rod  will  be  the 
difference  between  the  velocities  imparted  by  the  two  forces  ; 
and  if  these  be  laid  off  to  scale,  we  shall  have  the  second 
figure.  Hence  — 

A  pair  of  equal  and  opposite  parallel  forces,  applied  to  a 
straight  rod  of  uniform  section  and  material,  produce  a  rota- 
tion of  the  rod  about  its  centre.  Also,  — 

Such  a  rotation  about  the  centre  of  the  rod  cannot  be  pro- 


EFFECT  OF  STATICAL    COUPLE   ON  STRAIGHT  ROD.       51 

duced  by  a  single  force,  but  requires  a  pair  of  equal  and  op- 
posite parallel  forces. 

§  50.  Statical  Couple.  —  A  pair  of  equal  and  opposite 
parallel  forces  is  called  a  statical  couple. 

§51.  Effect  of  a  Single  Force  applied  at  the  Centre  of 
Gravity  of  a  Straight  Rod  of  Non-Uniform  Section  and 
Material.  —  In  the  case  of  a  straight  rod  of  non-uniform  sec- 
tion and  material,  we  may  consider  the  rod  as  composed  of  a 
set  of  particles  of  unequal  mass  :  and  if  we  imagine  each  par- 
ticle to  have  imparted  to  it  the  same  velocity  in  a  unit  of  time, 
then,  using  the  same  method  of  graphical  representation  as 
before  (Fig.  24),  the  line  ab,  bounding  the  other  ends  of  the 
lines  representing  velocities,  will  be  parallel  to  AB ;  but  if  we 
were  to  represent  by  the  lines  xy,  not  the  velocities  imparted, 
but  the  forces  per  unit  of  length,  the  line  bounding  the  other 
ends  of  these  forces  would  not,  in  this  case,  be  parallel  to  AB. 
Moreover,  since  these  forces  are  proportional  to  the  masses,  and 
hence  to  the  weights  of  the  several  particles,  their  resultant 
would  act  at  the  centre  of  gravity  of  the  rod.  Hence  — 

A  force  applied  at  the  centre  of  gravity  of  a  straight  rod  will 
impart  the  same  velocity  to  each  point  of  the  rod ;  i.e.,  will  im- 
part to  it  a  motion  of  translation  only. 

§  52.  Effect  of  a  Statical  Couple  on  a  Straight  Rod  of 
Non-Uniform  Section  and  Material.  —  Let  such  a  rod  have 
imparted  to  it  only  a  motion  of  rotation  about  its  centre  of 
gravity,  and  let  us  adopt  the  same  modes  of  graphical  repre- 
sentation as  before. 

Let  the  origin  be  taken  at  O  (Fig.  29), 
the  centre  of  gravity  of  the  rod. 

Let  Aa    =  ft  =  velocity  imparted  to  A. 
Bb    =  /2  =  velocity  imparted  to  B. 
OA  =  a,  OB  =  b,  OC  =  x. 
CD  =  f  =  velocity  imparted  to  C. 
dM  =  elementary  mass  at  C.  FIG.  29. 


52  APPLIED   MECHANICS. 

Then,  from  similar  triangles,  we  have 

/-i.-£         • 

a          b 
and  hence  for  the  force  acting  on  dM  we  have 

dF=(CE)dx   =  f±xdM. 


Hence  the  whole  force  acting  on  AO,  and  represented  graph- 
ically by  Aa,O,  is 

f 

J 


xdM, 
=  o 

and  that  acting  on  OB,  and  represented  by  B0blt  is 

f     (*x  =  o  /     f*x  =  o 

Jl  I    xdM  =  ^  I    xdM. 
vjx  =  -b  ajx  =  -b 

Hence  for  the  resultant,  or  the  algebraic  sum,  of  the  two,  we 
have 


R  =  *  f  Ok 

a  Jx  =  -b 


But  from  §  43  we  have  for  the  co-ordinate  x0  of  the  centre  of 
gravity  of  the  rod 

f*xdM 
_  Jx  =  -b     , 

M 
but,  since  the  origin  is  at  the  centre  of  gravity,  we  have 

X0  =   O, 

and  hence 

f*x  =  a 

o         /.     R  =  o. 


f*x  =  a 

I    xdM '  = 

Jx  =  -b 


Hence  the  two  forces  represented  by  Aa,O  and  Bb,O  are  equal 
in  magnitude  and  opposite  in  direction  :  hence  the  rotation 
about  the  centre  of  gravity  is  produced  by  a  Statical  Couple. 


MEASURE   OF   THE  ROTATORY  EFFECT.  53 

Now,  a  train  of  reasoning  similar  to  that  adopted  in  the  case 
of  a  rod  of  uniform  section  and  material  will  show  that  a  single 
force  applied  at  some  point  which  is  not  the  centre  of  gravity 
of  the  rod  will  produce  a  motion  which  consists  of  two  parts ; 
viz.,  a  motion  of  translation,  where  all  points  of  the  rod  have 
-equal  velocities,  and  a  motion  of  rotation  around  the  centre  of 
gravity  of  the  rod. 

§53.  Moment  of  a  Couple.  —  The  moment  of  a  statical 
couple  is  the  product  of  either  force  by  the  perpendicular  dis- 
tance between  the  two  forces,  this  perpendicular  distance  being 
called  the  arm  of  the  couple. 

§54.  Measure  of  the  Rotatory  Effect.  —  Before  proceed- 
ing to  examine  the  effect  of  a  statical  couple  upon  any  rigid 
body  whatever,  we  will  seek  a  means  of  measuring  its  effect  in 
the  cases  already  considered. 

The  measure  adopted  is  the  moment  of  the  couple ;  and,  in 
order  to  show  that  it  is  proper  to  adopt  this  measure,  it  will  be 
necessary  to  show  — 

That  the  moment  of  the  couple  is  proportional  to  the  angu- 
lar velocity  imparted  to  the  same  rod  in  a  unit  of  time ;  and 
from  this  it  will  follow  — 

That  two  couples  in  the  same  plane  with  equal  moments  will 
balance  each  other  if  one  is  right-handed  and  the  other  left-handed 

If  we  assume  the  origin  of  co-ordinates  at  C  (Fig.  30),  the 
centre  of  gravity  of  the  rod,  and  if  we 
denote  by  a  the  angular  velocity  imparted 
in  a  unit  of  time  by  the  forces  F  and  —F, 
.and  let  CD  =  x»  CE  =  x»  then  we  have 
for  the  linear  velocity  of  a  particle  situated 
at  a  distance  x  from  C  the  value 


ax.  FIG.  30. 

The  force  which  will  impart  this  velocity  in  a  unit  of  time  to 
the  mass  dM  is 


54  APPLIED  MECHANICS. 


The  total  resultant  force  is 


which,  as  we  have  seen,  is  equal  to  zero.     The  moment  of  the 
elementary  force  about  C  is 


and  the  sum  of  the  moments  for  the  whole  rod  is 


and  this,  as  is  evident  if  we  take  moments  about  C,  is  equal  t«, 
Fxi  -  Fx2  =  F(xs  -  xz)  =  F(DE). 

Now,  fx2dM  is  a  constant  for  the  same  rod  :  hence  any  quan- 
tity proportional  to  F(DE)  is  also  proportional  to  a. 

The  above  proves  the  proposition. 

Moreover,  we  have 

F(DE)  = 


whence  it  follows,  that  when  the  moment  of  the  couple  -is  givenr 
and  also  the  rod,  we  can  find  the  angular  velocity  imparted  in 
a  unit  of  time  by  dividing  the  former  by  fx*dM. 

§  55.   Effect   of   a  Couple  on  a  Straight  Rod  when   the 

Forces  are  inclined  to  the  Rod.  —  We  shall  next  show  that 

the  effect  of  such  a  couple  is  the  same  as  that  of  a  couple  of 

equal  moment  whose  forces  are  perpen- 

f""r;^Ir'^  dicular  to  the  rod. 

I/I        ^^.^  In  this  case  let  AD  and  BC  be  the 

'    ^8   forces  (Fig.  ^i}.      The   moment    of   this 
couple  is  the  product  of  AD  by  the  per- 
c         '  pendicular  distance  between  AD  and  BC, 
the  graphical  representation  of  this  being 
the  area  of  the  parallelogram  ADBC. 


EFFECT  OF  A   STATICAL    COUPLE    ON  A   RIGID   BODY.    55 

Resolve  the  two  forces  into  components  along  and  at  right 
angles  to  the  rod.  The  former  have  no  effect  upon  the  motion 
of  the  rod  :  the  latter  are  the  only  ones  that  have  any  effect 
upon  its  motion.  The  moment  of  the  couple  which  they  form 
is  the  product  of  Ad  by  AB,  graphically  represented  by  paral- 
lelogram AdBb ;  and  we  can  readily  show  that 

ADBC  =  AdBb. 

Hence  follows  the  proposition. 

§  56.  Effect  of  a  Statical  Couple  on  any  Rigid  Body.  — 

Refer  the  body  (Fig.  32)  to  three  rectan- 
gular axes,  OX,  OY,  and  OZ,  assuming 
the  origin  at  the  centre  of  gravity  of  the 
body,  and  OZ  as  the  axis  about  which 
the  body  is  rotating.  Let  the  mass  of  the 
particle  P  be  AJ/,  and  its  co-ordinates  be 
v>  7,  z. 

Then  will  the  force  that  would  impart  FIG.  32. 

to  the  mass  &M  the  angular  velocity  a  in  a  unit  of  time  be 


where  r  =  perpendicular  from  P  on  OZ,  or 

r  =  V^2  -h  jv2. 

This  force  may  be  resolved  into  two,  one  parallel  to  OY  and 
the  other  to  OX ' ;  the  first  component  being  &x&M,  and  the 
second  ay^M. 

Proceeding  in  the  same  way  with  each  particle,  and  finding 
the  resultant  of  each  of  these  two  sets  of  parallel  forces,  we 
shall  obtain,  finally,  a  single  force  parallel  to  OY  and  equal  to 


and  another  parallel  to  OX,  equal  to 


56 


APPLIED   MECHANICS. 


But,  since  OZ  passes  through  the  centre  of  gravity  of  the  body, 
we  shall  have 

=  o     and 


—  o. 


Hence  the  resultant  is  in  each  case,  not  a  single  force,  but  a 
statical  couple  ;  the  moment  of  the  first  couple  being 


and  that  of  the  second 


These  couples  produce  the  same  effect  in  whatever  plane  per- 
pendicular to  OZ  they  are  situated.  Hence,  suppose  them  both 
in  the  plane  XO  Y,  then  representing  them  as  in  (Fig.  33),  we 
make 

F,(AB)  =  o^x^M    and     F±(CD)  =  a2 


Now,  compound  the  force  at  D  with  that  at  B,  and  the  force  at 

t_^ ,  C  with  that  at  A,  and  we  obtain  as  a 

result  two  equal  and  opposite  parallel 

x.    forces,   or  one  statical  couple,   whose 

K        moment  will   be   shown   to   be   equal 


-T 


to 


To  show  this,  let 

OB  —  pl  =  one-half  the  arm  of  a 
OD  —  p2  =  one-half  the  arm  of  a' 
Oe  —  p  =  one-half  the  arm  of  the  resultant  couple. 

Let  angle  DOe  —  0  =  GEK:  we  shall  then  have 
/  =  Oe  =  Of  +  fe  =  /a  cos  0  4-  A  sin  0  ; 


COUPLES  IN  THE  SAME   OR  PARALLEL   PLANES. 
but 

...  p 

.'  R          R 

.-.    Rp     =  F^p,  +  F2pz. 

Hence  the  moment  of  the  resultant  couple  is  equal  to  the  sum 
of  the  moments  of  the  separate  couples,  or 

R(2p)  =  o^r^M  =  a%x2&M  +  a2y2&M. 

Hence  :  To  impart  to  a  body  a  rotation  about  an  axis  passing 
through  its  centre  of  gravity  requires  the  action  of  a  statical 
couple,  and  conversely  a  statical  couple  so  applied  will  cause  such 
a  rotation  as  that  described. 

Hence  we  may  generalize  all  our  propositions  in  regard  to 
the  effect  of  statical  couples  and  we  may  conclude  that  — 

In  order  that  two  couples  may  have  the  same  effect,  it  is 
necessary  — 

i°.    That  they  be  in  the  same  or  parallel  planes. 

2°.    That  they  have  the  same  moment. 

3°.  That  they  tend  -to  cause  rotation  in  the  same  direction 
{i.e.,  both  right-handed  or  both  left-handed  when  looked  at  from 
the  same  side]. 

It  also  follows,  that,  for  a  given  statical  couple,  we  may  sub- 
stitute another  having  the  magnitudes  of  its  forces  different, 
provided  only  the  moment  of  the  couple  remains  the  same. 

§  57.  Composition  of  Couples  in  the  Same  or  Parallel 
Planes.  —  If  the  forces  of  the  couples  are  not 
the  same,  reduce  them  to  equivalent  couples 
having  the  same  force,  transfer  them  to  the 
same  plane,  and  turn  them  so  that  their  arms 
shall  lie  in  the  same  straight  line,  as  in  Fig. 
34;  the  first  couple  consisting  of  the  force  F  FlG'34' 

at  A  and  —  F  at  B,  and  the  second  of  F  at  B  and  —  F  at  C. 


APPLIED   MECHANICS. 


The  two  equal  and  opposite  forces  counterbalance  each  other, 
and  we  have  left  a  couple  with  force  F  and  arm 

AC  =  AB  +  BC 
:.    Resultant  moment  =  F.  AC  =  F(AB)  -f  F(J3C). 

Hence :  The  moment  of  the  couple  which  is  the  resultant  of 
two  or  more  couples  in  the  same  or  parallel  planes  is  equal  to 
the  algebraic  sum  of  the  moments  of  the  component  couples. 

EXAMPLES. 

i.  Convert  a  couple  whose  force  is  5  and  arm  6  to  an  equivalent 
couple  whose  arm  is  3.  Find  the  resultant  of  this  and  another  couple 
in  the  same  plane  and  sense  whose  force  is  7  and  arm  8 ;  also  find  the 
force  of  the  resultant  couple  when  the  arm  is  taken  as  5. 


Solution. 


Moment  of  first  couple  =    5  x     6  =  30 

When  arm  is  3,  force  =    -^  =  10 

Moment  of  second  couple  =  7  x  8  =  56 
Moment  of  resultant  couple  =  30  +  56  =  86 
When  arm  is  5,  force  =  -8-6-  =  X7i 

2.  Given  the  following  couples  in  one  plane  :  — 


Force. 

Arm. 

12 

17  1 

3 

8 

5 

7 

6 

9 

12 

12 

IO 

9 

14 

6  J 

Convert  to  equivalent 
couples  having  the 
following :  — 


Force. 

5 

8 
6 


Arm. 


20 


The  first  and  the  last  three  are  right-handed ;  the  second,  third,  and 
fourth  are  left-handed.  Find  the  moment  of  the  resultant  couple,  and 
also  its  force  when  it  has  an  arm  n. 


COUPLES  IN  PLANES  INCLINED    TO   EACH  OTHER.       59 

§58.  Representation  of  a  Couple  by  a  Line.  —  From  the 
preceding  we  see  that  the  effect  of  a  couple  remains  the  same 
as  long  as  — 

I  °.  Its  moment  does  not  change. 

2°.  The  direction  of  its  axis  (i.e.,  of  the  line  drawn  perpen- 
dicular to  the  plane  of  the  couple]  does  not  change. 

3°.  The  direction  in  which  it  tends  to  make  the  body  turn 
(right-handed  or  left-handed}  remains  the  same. 

Hence  a  couple  may  be  represented  by  drawing  a  line  in 
the  direction  of  its  axis  (perpendicular  to  its  plane),  and  laying 
off  on  this  line  a  distance  containing  as  many  units  of  length 
as  there  are  units  of  moment  in  the  couple,  and  indicating  by  a 
dot,  an  arrow-head,  or  some  other  means,  in  what  direction  one 
must  look  along  the  line  in  order  that  the  rotation  may  appear 
right-handed. 

This  line  is  called  the  Moment  Axis  of  the  couple. 

§  59.  Composition  of  Couples  situated  in  Planes  inclined 
to  Each  Other.  —  Suppose  we  have  two  couples  situated 
neither  in  the  same  plane  nor  in  parallel  planes,  and  that  we 
wish  to  find  their  resultant  couple.  We  may  proceed  as  fol- 
lows :  Substitute  for  them  equivalent  couples  with  equal  arms, 
then  transfer  them  in  their  own  plane  respectively  to  such  posi- 
tions that  their  arms  shall  _Rt 
coincide,  and  lie  in  the 
line  of  intersection  of  the 
two  planes. 

This  having  been  done, 
let  OOt  (Fig.  35)  be  the 
common  arm,  F  and  —  F 
the  forces  of  one  couple, 
Fl  and  —Fl  those  of  the 
other.  The  forces  F  and 
Fs  have  for  their  resultant  R,  and  —  F  and  —F,  have  —  Rlt 
Moreover,  we  may  readily  show  that  R  and  —  R1  are  equal  and 


60  APPLIED   MECHANICS. 

parallel,  both  being  perpendicular  to  OOlt  The  resultant  of 
the  two  couples  is,  therefore,  a  couple  whose  arm  is  OO^  and 
force  R,  the  diagonal  of  the  parallelogram  on  F  and  Flt  so  that 


R  =  \F*  +  jr*  +  2FFt  cos  0, 

where  0  is  the  angle  between  the  planes  of  the  couples.  Now, 
if  we  draw  from  O  the  line  Oa  perpendicular  to  OOl  and  to  F, 
and  hence  perpendicular  to  the  plane  of  the  first  couple,  and  if 
we  draw  in  the  same  manner  Ob  perpendicular  to  the  plane  of 
the  second  couple,  so  that  there  shall  be  in  Oa  as  many  units 
of  length  as  there  are  units  of  moment  in  the  first  couple,  and 
in  Ob  as  many  units  of  length  as  there  are  units  of  moment  in 
the  second  couple,  we  shall  have  — 

i°.  The  lines  Oa  and  Ob  are  the  moment  axes  of  the  two 
given  couples  respectively. 

2°.  The  lines  Oa  and  Ob  lie  in  the  same  plane  with  F  and 
FIt  this  plane  being  perpendicular  to  OOt. 

3°.  We  have  the  proportion 

Oa-.Ob  =  F.  OO, :  F, .  OO,  =  F :  F,. 

4°.  If  on  Oa  and  Ob  as  sides  we  construct  a  parallelogram, 
it  will  be  similar  to  the  parallelogram  on  F  and  Ft.  We  shall 
have  the  proportion 

Oc-.R=  Oa:F  =  Ob-.F,; 

and  since  the  sides  of  the  two  parallelograms  are  respectively 
perpendicular  to  each  other,  the  diagorials  are  perpendicular  to 
each  other ;  and  since  we  have  also 

Oc  =  R  '  Oa     and     Oa  =  F.  OOt        .'.     Oc  =  R  .  OOIt 

it  follows  that  Oc  is  perpendicular  to  the  plane  of  the  resultant 
couple,  and  contains  as  many  units  of  length  as  there  are  units 
of  moment  in  the  moment  of  the  resultant  couple;  in  other 


COUPLE  AND   SINGLE  FORCE  IN  THE  SAME  PLANE.     6 1 

words,  Oc  will  represent    the   moment   axis   of   the   resultant 
couple,  and  we  shall  have 


Oc  =  \Oa2  +  Ob2  +  2Oa  .  ObcosaOb; 
or,  if  we  let 

Oa  =  Z,     Ob  =  M,     Oc  =  G,     aOb  =  0, 


G  =  \lL2  +  M2  +  2ZJ/cos  6. 

This  determines  the  moment  of  the  resultant  couple ;  and,  for 
the  direction  of  its  moment  axis,  we  have 


and 


sin  a  Of  =  —sinO 
G 


smbOc  =  —  sin0. 

Cr 


Hence  we  can   compound  and  resolve  couples  just  as  we  do 
forces,  provided  we  represent  the  couples  by  their  moment  axes 

EXAMPLES.   . 

1.  Given     L  =  43,     J/=  15,     6  =  65°;  find  resultant  couple. 

2.  Given     Z  =  40,    -M  =  30,     6  =  30°  ;  find  resultant  couple. 

3.  Given     Z  =     i,     M  =     5,     0  =  45°;  find  resultant  couple. 

§  60.   Resultant  of  a  Couple  and  a  Single  Force  in  the 
Same  Plane Let  M  (Fig.  36  or  37)  be  the  moment  of  the 


FIG.  36. 


given  couple,  and  let  OF  —  F  be  the  single  force.  For  the 
given  couple  substitute  an  equivalent  couple,  one  of  whose 
forces  is  —  F  at  O,  equal  and  directly  opposed  to  the  single 


62 


APPLIED   MECHANICS. 


force  F,  these  two  counterbalancing  each  other,  and  leaving 
only  the  other  force  of  the  couple,  which  is  equal  and  parallel 
to  the  original  single  force  F,  and  acts  along  a  line  whose 


distance  from  O  is  OA  =      . 

F 


Hence  — 


The  resultant  of  a  single  force  and  a  couple  in  the  same  plane 
is  a  force  equal  and  parallel  to  the  original  force,  having  its 
line  of  direction  at  a  perpendicular  distance  from  the  original 
force  equal  to  the  moment  of  the  couple  divided  by  the  force. 

Fig.  36  shows  the  case  when  the  couple  is  right-handed,  and 
Fig.  37  when  it  is  left-handed. 

§61.  Composition  of  Parallel  Forces  in  General.  —  In 
each  case  of  composition  of  parallel  forces  (§§  34,  35,  and  36) 
it  was  stated  that  the  method  pursued  was  applicable  to  all 
cases  except  those  where 


=  o. 


We  were  obliged,  at  that  time,  to  reserve  this  case,  because  we 
had  not  studied  the  action  of  a  statical  couple ;  but  now  we  will 
adopt  a  method  for  the  composition  of 'parallel  forces  which  will 
apply  in  all  cases. 

(a)  When  all  the  forces  are  in  one  plane.  Assume,  as  we  did 
in  §35,  the  axis  OY  to  be  parallel  to 
the  forces ;  assume  the  forces  and  the 
co-ordinates  of  their  lines  of  direction, 
as  shown  in  the  figure  (Fig.  38).  Now 
place  at  the  origin  O,  along  OY,  two 
equal  and  opposite  forces,  each  equal  to 
x  Ft ;  then  these  three  forces,  viz.,  Ft  at 
D,  OA,  and  OB,  produce  the  same  effect 
as  F,  at  D  alone ;  but  F,  at  D  and  OB 

B*  form  a  couple  (left-handed  in  the  figure) 

FlG<38-  whose   moment   is  — /*>,.      Hence   the 

force  FL  is  equivalent  to  — 


COMPOSITION  OF  PARALLEL   FORCES. 


i°.  An  equal  and  parallel  force  at  the  origin,  and 

2°.  A  statical  couple  who'se  moment  is  —  /v*1,. 

Likewise  the  force  F2  is  equivalent  to  (i°)  an  equal  and  par- 
allel force  at  the  origin,  and  (2°)  a  couple  whose  moment  is 
—  F2x2,  etc. 

Hence  we  shall  have,  if  we  proceed  in  the  same  way  with 
all  the  forces,  for  resultant  of  the  entire  system  a  single  force 


R  = 


along  OY, 


and  a  single  resultant  couple 

M=  - 

(Observe  that  downward  forces  and  left-handed  couples  are 
to  be  accounted  negative.) 

Now,  there  may  arise  two  cases. 
i°.  When  ^F  '=  o,  and 
2°.  When  $FXo. 

CASE  I.  When  2F  =  o,  the  resultant  force  along  0  Y  van- 
ishes, and  the  resultant  of  the  entire  system  is 
a  statical  couple  whose  moment  is 


CASE  II.  When  %F><o,  we  can  reduce 
the  resultant  to  a  single  force. 

Let  (Fig.  39)  OB  represent  the  resultant 
force  along  O  Y,  R  =  %F.  With  this,  compound 
the  couple  whose  moment  is  M  =  —?<Fx,  and 
we  obtain  as  resultant  (§  60)  a  single  force 

R  =  ^F, 


FIG.  39. 


whose  line  of  action  is  at  a  perpendicular  distance  from  OY 
equal  to 

AO 


ocr 


64 


APPLIED   MECHANICS. 


(b]    When  the  forces  are  not  confined  to  one  plane.     Assume, 
as  before  (Fig.  40),  OZ  parallel  to  the  forces,  and  let  F  acting 

through  A  be  one  of  the  given 
forces,  the  co-ordinates  of  A  be- 
ing x  and  y.  Place  at  O  two  equal 
and  opposite  forces,  each  equal  to 
F,  and  also  at  B  two  equal  and 
opposite  forces,  each  equal  to  F. 
These  five  forces  produce  the 
same  effect  as  F  alone  at  A,  and 
they  may  be  considered  to  con- 
sist of  — 


FIG.  40. 


i°.  A  single  force  F  at  the  origin. 


and 


2°.  A  couple  whose  forces  are  F  at  B  and  —  F  at 
whose  moment  is  —  Fx  acting  in  the  y  plane. 

3°.  A  couple  whose  forces  are  F  at  A  and  —  F  at  B,  and 
whose  moment  is  Fy  acting  in  the  x  plane.  Treating  each  of 
the  forces  in  the  same  way,  we  shall  have,  in  place  of  the  entire 
system  of  parallel  forces,  the  following  forces  and  couples  :  — 

i°.  A  single  force  R  —  2F  along  OZ. 

2°.  A  couple  My  —  —  ^Fx  in  the  y  plane. 

3°.  A  couple  Mx  —  -\-^Fy  in  the  x  plane. 

Now,  there  may  be 
two  cases  :  —  H 

i°.  When  2F  ><  o. 


R=SF 


2°.  When 


=  o. 


CASE  I.  When 
<  o,  we  can  reduce  to  a 
single  resultant  force 
having  a  fixed  line  of 
direction.  Lay  off  (Fig. 
41  )  along  OZ,  OH  =  ^F. 
Combining  this  with  the  first  of  the  above-stated  couples,  we 


FIG.  41. 


COMPOSITION  OF  PARALLEL   FORCES. 


obtain  a  force  R  =  2,F  at  A,  where  OA  =  -  = 


Thea 


combine  with  this  resultant  force  R  =.  2F  at  A,  the  second 
couple,  and  we  shall  have  as  single  resultant  of  the  entire 
system  a  single  force 

R  =  2F 
acting  through  B,  where 


Hence  the  resultant  is  a  force  whose  magnitude  is 


the  co-ordinates  of  its  line  of  direction  being 


CASE  II.  When  3F=  o,  there  is  no  single  resultant  force ; 
but  the  system  reduces  to  two  couples,  one  in  the  x  plane  and 
one  in  the  y  plane,  and  these  two  can  be  reduced  to  one  single 
resultant  couple.  (Observe  that  couples  are  to  be  accounted 
positive  when,  on  being  looked  at  by  the  observer  from  the  posi- 
tive part  of  the  axis  towards 
the  origin,  they  are  right- 
handed  ;  otherwise  they  are 
negative.) 

The  moment  axis  of  the 
couple  in  the  x  plane  will 
be  laid  off  on  the  axis  OX 
from  the  origin  towards  the 
positive  side  if  the  moment 
is  positive,  or  towards  the 
negative  side  if  it  is  nega-  FIG.  42. 

tive,  and  likewise  for  the  couple  in  the  y  plane. 


66 


APPLIED   MECHANICS. 


Hence  lay  off  (Fig.  42)  OB  =  Mx,  OA  =  My,  and  by 
completing  the  rectangle  we  shall  have  OD  as  the  moment 
axis  of  the  resultant  couple ;  hence  the  resultant  couple  lies 
in  a  plane  perpendicular  to  OD,  and  its  moment  bears  to 
OD  the  same  ratio  as  Mx  bears  to  QM. 

Hence  we  may  write 

OD  =  Mr  =  V^/*2  +  Mf, 
cosDOX    =  ^  =  cose. 

Mr 

If  My  had  been  negative,  we  should  have  OE  as  the  moment 
axis  of  the  resultant  couple. 


EXAMPLES. 


F. 

X. 

y> 

F. 

X. 

j. 

I. 

5 

4 

3 

2. 

5 

—4 

3 

3 

2 

i 

—  2 

2 

—  i 

i 

3 

5 

-3 

3 

5 

Find  the  resultant  in  each  example. 

§62.  Resultant  of  any  System  of  Forces  acting  at  Dif- 
ferent Points  of  a  Rigid 
Body,  all  situated  in  One 
Plane.  —  Let  CF  =  F  (Fig. 
43)  be  one  of  the  given  forces. 
Let  all  the  forces  be  referred 
to  a  system  of  rectangular 
_x  axes,  as  in  the  figure,  and  let 
a  =  angle  made  by  F  with 
OX,  etc.  Let  the  co-ordi- 
nates of  the  point  of  application  of  Fbe  AO  =  x,  BO  =  y. 


M      A 


FIG.  43- 


SYSTEM  OF  FORCES  ACTING   ON  RIGID   BODY.  6/ 

We  first  decompose  CF  =  F  into  two  components,  parallel 
respectively  to  OX  and  O  Y.  These  components  are 

CD  =  Fcosa,     CE  —  Fsina. 

Apply  at  O  in  the  line  O  Y  two  equal  and  opposite  forces,  each 
equal  to  .Fsin  a,  and  at  O  in  the  line  OX  two  equal  and  opposite 
forces,  each  equal  to  Fcosa.  Since  these  four  are  mutually 
balanced,  they  do  not  alter  the  effect  of  the  single  force ;  and 
hence  we  have,  in  place  of  F  at  C,  the  six  forces  CD,  OM,  OK, 
CE,  ON,  OG.  Of  these  six,  CE  and  OG  form  a  couple  whose 
moment  is 

—  (Fsma)x  =  —Fxsma, 

CD  and  OK  form  a  couple  whose  moment  is 
(Fcosa)y  =  lycos  a. 

These  two  couples,  being  in  the  same  plane,  give  as  result- 
ant moment  their  algebraic  sum,  or 

F(y  cos  a  —  x  sin  a) . 

We  have,  therefore,  instead  of  the  single  force  at  Ct  the  follow- 
ing:— 

i°.   OM  =  Fcosa  along  OX. 

2°.   ON  =  Fsin  a  along  OY. 

3°.  The  couple  M  =  F(y  cos  a  —  x  sin  a)  in  the  given  plane. 

Decompose  in  the  same  way  each  of  the  given  forces ;  and 
we  have,  on  uniting  the  components  along  OX,  those  along  OY, 
and  the  statical  couples  respectively,  the  following :  — 

i°.  A  resultant  force  along  OX,  Rx  =  %Fcos  a. 

2°.  A  resultant  force  along  OY,  Ry  =  ^Fsina. 

3°.  A  resultant  couple  in  the  plane,  whose  moment  is 

M  =  2F(y  cos  a  —  x  sin  a) . 


68 


APPLIED   MECHANICS. 


This   entire    system,   on    compounding   the    two  forces    at   O, 
reduces  to 


sin  a)2; 


_O  75     */  Z>     -       I         Z?    •» 

I  .         K  =  \  fi-x2  -f-  Kv*  = 

*r 

making  with  6Uf  an  angle  <v,  where 
cosar  = 

K 

2°.  A  resultant  couple  in  the  same  plane,  whose  moment  is 
M  ==  *2<F(y  cos  a  —  x  sin  a) . 


Now  compound  this  resultant  force  and  couple,  and  we  have, 

for  final  resultant,  a  single 
force  equal  and  parallel  to 

E  R,  and  acting  along  a  line 
whose  perpendicular  dis- 

D     tance  from  O  is  equal  to 

M 


FIG.  44. 

Suppose  (Fig.  44)  the  force 

OB  =  SFcosa, 
OA  —  ^F  sin  a, 


x    The  equation  of   this  line 
may  be  found  as  follows  : 


cos  a)2  + 

and  let  us  suppose  the  resultant  couple  to  be  right-handed,  and 
let 


then  will  the  line  ME  parallel  to  OR  be  the  line  of  direction 
of  the  single  resultant  force. 


CONDITIONS  OF  EQUILIBRIUM.  69 

Assuming  the  force  R  to  act  at  any  point  C  (xrj  jj/r)  of  this 
line,  if  we  decompose  it  in  the  same  way  as  we  did  the  single 
forces  previously,  we  obtain  — 

i°.  The  force  R  cosar  =  ^Fcosa  along  OX. 

2°.  The  force  R  sin  ar  =  5LFsin  a  along  OY. 

3°.  The  couple  R(yr  cos  ar  —  xr  sin  c^). 

Hence  we  must  have 

jR(yr  cos  ar  —  xr  sin  ar)  =  *2<F(y  cos  a  —  #  sin  a)  =  M.   ,  -. 


Hence  for  the  equation  of  the  line  of  direction  we  have 

M  (  \ 

yr  cos  ar  —  xr  sin  ar  —  —  -.  (i) 

K 

Another  form  for  the  same  equation  is 

yr(%Fcosa)  —  xr(2Fsma)  =  M.  (2) 

§  63.  Conditions  of  Equilibrium.  —  If  such  a  set  of  forces 
be  in  equilibrium,  there  must  evidently  be  no  tendency  to  trans- 
lation and  none  to  rotation.  Hence  we  must  have 

R  =  o     and     M  =  o. 

Hence  the  conditions  of  equilibrium  for  any  system  of  forces 
in  a  plane  are  three  ;  viz.,  — 


==  o,     ^F  sin  a  =  o,     ^(jycosa  —  #sma)  =  o. 

Another  and  a  very  convenient  way  to  state  the  conditions  of 
•equilibrium  for  this  case  is  as  follows  :  — 

If  the  forces  be  resolved  into  components  along  two  directions 
at  right  angles  to  each  other,  then  the  algebraic  sum  of  the  com- 
ponents along  each  of  these  directions  must  be  zero,  and  the 
algebraic  sum  of  the  moments  of  the  forces  about  any  axis  per~ 
pendicular  to  the  plane  of  the  forces  must  equal  zero. 


APPLIED   MECHANICS. 


EXAMPLES. 


I.  Given 


2.  Given 


P. 

X. 

* 

a. 

5 

3 

2 

31° 

10 

i 

3 

40° 

-7 

4 

2 

54° 

F. 

X. 

* 

a. 

12 

27 

3 

15° 

4 

T3 

-5 

30° 

8 

-5 

—4 

45° 

Find  the  resultant,  and 
the  equation  of  its 
line  of  direction. 

15°  "I  Find  the  resultant,  and 
y  the  equation  of  its 
J  line  of  direction. 


§64.  Resultant  of  any  System   of   Forces   not  confined 

to  One  Plane Suppose  we 

have  a  number  of  forces  applied 
at  different  points  of  a  rigid 
body,  and  acting  in  different 
directions,  of  which  we  wish  to- 
find  the  resultant.  Refer  them 
all  to  a  system  of  three  rect- 
x  angular  axes,  OX,  OY,  OZ 
(Fig.  45).  Let  PR  —  F  be 
one  of  the  given  forces.  Re- 
solve it  into  three  components, 
PK,  PH,  and  PG,  parallel 
Let 


FIG.  45- 

respectively  to  the  three  axes. 


RPK  =  a,     RPH  =  J3,     RPG  =  y. 

Let  OA  =  x,  OB  =  y,  OC  —  2,  be  the  co-ordinates  of  the 
point  of  application  of  the  force  F.  Now  introduce  at  B  and 
also  at  O  two  forces,  opposite  in  direction,  and  each  equal  to  PK, 
We  now  have,  instead  of  the  force  PK,  the  five  forces  PK,  BM, 
BN,  OS,  and  OT.  The  two  forces  PK  and  BN  form  a  couple 
in  the  y  plane,  whose  axis  is  a  line  parallel  to  the  axis  O  Y,  and 
whose  moment  is  (PK)(EB}  —  (Fcosa)z  =  Fzcosa.  The 


FORCES  NOT  CONFINED    TO   ONE   PLANE.  ?l 

forces  BM  and  OT  form  a  couple  in  the  z  plane,  whose  moment 

is 

(BM)(OB)  =  -Jycosa. 

Now  do  the  same  for  the  other  forces  PH  and  PG,  and  we  shall 
finally  have,  instead  of  the  force  PR,  three  forces, 

F  cos  a,    F  cos  ft,     F  cos  y, 

acting  at  (9  in  the  directions  OX,  OY,  and  OZ  respectively, 
together  with  six  couples,  two  of  which  are  in  the  x  plane,  two 
in  the  y  plane,  and  two  in  the  z  plane. 

They  thus  form  three  couples,  whose  moments  are  as  fol- 
lows :  — 

Around  OX,  F(y  cos  y  —  z  cos  ft) ; 

Around  OY,  F(z  cos  a  —  x  cos  y) ; 

Around  OZ,  F(x  cos  ft  —y  cos  a) . 

Treat  each  of  the  given  forces  in  the  same  way,  and  we  shall 
have,  in  place  of  all  the  forces  of  the  system,  three  forces, 

^F  cos  a  along  OX, 
along  OY, 
along  OZ; 


and  three  couples,  whose  moments  are  as  follows :  — 

Around  OX,  Mx  =  ^F(y  cos  y  —  z  cos  ft) ; 
Around  O  Y,  My  =  *%F(z  cos  a  —  x  cos  y) ; 
Around  OZ,  Mz  ==  ^(x  cos  ft  —  jycosa). 

The  three  forces  give  a  resultant  at  O  equal  to 
R  =  V(2^cosa)2  +  (XFcos/3)2  + 

COS  Or  =  -,        COS  ftr  =  ^-,       COS  yr 

R  R 


APPLIED   MECHANICS. 


For  the  three  couples  we 

«V-i 

have  as  resultant 

(3) 

(4) 

\lMx2  +  My2  -h  J/22, 

cos/x  =  ^,       cosv  =  ^; 

n         ^//~?                      ^/f  ' 

A,  /x,,  and  v  being  the  angles  made  by  the  moment  axis  of  the 
resultant  couple  with  OX,  O  Y,  and  OZ  respectively. 

Thus  far  we  have  reduced  the  whole  system  to  a  single  result- 
ant force  at  the  origin,  and  a  couple.  Sometimes  we  can  reduce 

the  system  still  farther 
and  sometimes  not.  The 
following  investigation  will 
show  when  we  can  do  so. 
Let  (Fig.  46)  OP  —  R  be 
the  resultant  force,  and 
OC  =M  the  moment  axis 
of  the  resultant  couple. 
Denote  the  angle  between 
them  by  0  (a  quantity  thus 
far  undetermined).  Pro- 
ject OP  =  R  on  OC.  Its 
projection  will  be  OD  =  RcosO;  then  project,  in  its  stead,  the 
broken  line  OABP  on  OC.  By  the  principles  of  projections, 
the  projection  of  this  broken  line  will  equal  OD. 

Now  OA,  AB,  and  BP  are  the  co-ordinates  of  P,  and  make 
with  OC  the  same  angle  as  the  axes  OX,  OY,  and  OZ  '  ;  i.e., 
A,  p,  and  v  respectively  :  hence  the  length  of  the  projection  is 


PIG.  46. 


OA  cos  A  -|-  AB  cos  /*  +  BPcos  v. 


But 


Hence 


OA  =  /?  cosar,     AB  = 


r,     BP  =  ficosyr. 


ItcosO  =  7?  cos  ar  cos  A  -J-  />  cos  /2r  cos  /A  +  ^cosyrcos»/ 
cos0      =  cos  OT  cos  A      -f  cos  fir  cos  /x,      +  cosyrcosv.       (5) 


CONDITIONS   OF  EQUILIBRIUM.  73 

This  enables  us  to  find  the  angle  between  the  resultant  force 
and  the  moment  axis  of  the  resultant  couple. 

The  following  cases  may  arise  :  — 

i°.  When  cos  6  =  o,  or  6  =  90°,  the  force  lies  in  the  plane 
of  the  couple,  and  we  can  reduce  to  a  single  force  acting  at  a 

M 

distance  from   O  equal  to  —  ,  and  parallel  to  R  at  O\ 

R 

2°.  When  cos0  =  i,  or  0  =  o,  the  moment  axis  of  the 
couple  coincides  in  direction  with  the  force  :  hence  the  plane 
of  the  couple  is  perpendicular  to  the  force,  and  no  farther 
reduction  is  possible. 

3°.  When  0  is  neither  o°  nor  90°,  we  can  resolve  the  couple 
M  into  two  component  couples,  one  of  which,  JWcosO,  acts  in  a 
plane  perpendicular  to  the  direction  of  R,  and  the  other,  Msm  6, 
acts  in  a  plane  containing  R.  The  latter,  on  being  combined 
with  the  force  R  at  the  origin,  gives  an  equal  and  parallel  force 
whose  line  of  action  is  at  a  distance  from  that  of  R  at  O,  equal 
to 

MsmO 


4°.  When  M  =  o,  the  resultant  is  a  single  force  at  O. 

5°.  When  R  =  o,  the  resultant  is  a  couple. 

§65.  Conditions  of  Equilibrium.  —  To  produce  equilibrium, 
we  must  have  no  tendency  to  translation  and  none  to  rotation. 
Hence  we  must  have 

R  =  o     and    M  =  o. 
Hence  we  have,  in  general,  six  conditions  of  equilibrium  ;  viz.,  — 


2F  cos  a  =  o,     'SFcosfi  =  o, 

Mx  =    O,        My  =    O,        MZ 


74  APPLIED  MECHANICS. 


EXAMPLES. 

1.  Prove  that,  whenever  three  forces  balance  each  other,  they  must 
lie  in  one  plane. 

2.  Show  how  to  resolve  a  given  force  into  two  whose  sum  is  given,, 
the  direction  of  one  being  also  given. 

3.  A  straight  rod  of  uniform  section  and  material  is  suspended  by  two 
strings  attached  to  its  ends,  the  strings  being  of  given  length,  and  attached 
to  the  same  fixed  point :  find  the  position  of  equilibrium  of  the  rod. 

4.  Two  spheres  are  supported  by  strings  attached  to  a  given  point, 
and  rest  against  each  other :  find  the  tensions  of  the  strings. 

5 .  A  straight  rod  of  uniform  section  and  material  has  its  ends  resting 
against  two  inclined  planes  at  right  angles  to  each  other,  the  vertical 
plane  which  passes  through  the  rod  being  at  right  angles  to  the  line  of 
intersection  of  the  two  planes  :  find  the  position  of  equilibrium  of  the 
rod,  and  the  pressure  on  each  plane,  disregarding  friction. 

6.  A  certain  body  weighs  8  Ibs.  when  placed  in  one  pan  of  a  false 
balance  of  equal  arms,  and  10  Ibs.  in  the  other :  find  the  true  weight  of 
the  body. 

7.  The  points  of  attachment  of  the  three  legs  of  a  three-legged  table 
are  the  vertices  of  an  isosceles  right-angled  triangle  ;  a  weight  of  100  Ibs. 
is  supported  at  the  middle  of  a  line  joining  the  vertex  of  one  of  the  acute 
angles  with  the  middle  of  the  opposite  side :    find  the  pressure  upon 
each  leg. 

8.  A  heavy  body  rests  upon  an  inclined  plane  without  friction  :  find 
the  horizontal  force  necessary  to  apply,  to  prevent  it  from  falling. 

9.  A  rectangular  picture  is  supported   by  a  string   passing  over  a 
smooth  peg,  the  string  being  attached  in  the  usual  way  at  the  sides,  but 
one-fourth  the  distance  from  the  top  :  find  how  many  and  what  are  the 
positions  of  equilibrium,  assuming  the  absence  of  friction. 

10.  Two  equal  and  weightless  rods  are  jointed  together,  and  form  a 
right  angle ;   they  move   freely  about  their  common    point :    find   the 
ratio  of  the  weights  that  must  be  suspended  from  their  extremities,  that 
one  of  them  may  be  inclined  to  the  horizon  at  sixty  degrees. 

11.  A  weight  of  100  Ibs.  is  suspended  by  two  flexible  strings,  one 
of  which  is  horizontal,  and  the  other  is  inclined  at  an  angle  of  thirty 
degrees  to  the  vertical :  find  the  tension  in  each  string. 


D  YNAMICS.  —  DEFINITIONS.  7  5 


CHAPTER  II. 

DYNAMICS. 

§  66.  Definitions.  —  Dynamics  is  that  part  of  mechanics 
which  discusses  the  forces  acting,  when  motion  is  the  result. 

Velocity,  in  the  case  of  uniform  motion,  is  the  space  passed 
over  by  the  moving  body  in  a  unit  of  time  ;  so  that,  if  s  repre- 
sent the  space  passed  over  in  time  t,  and  v  represent  the  velocity, 
then 


Velocity,  in  variable  motion,  is  the  limit  of  the  ratio  of  the 
space  (AJ)  passed  over  in  a  short  time  (A^),  to  the  time,  as  the 
latter  approaches  zero  :  hence 


Acceleration  is  the  limit  of  the  ratio  of  the  velocity  (A^)  im- 
parted to  the  moving  body  in  a  short  time  (A/),  to  the  time,  as 
the  time  approaches  zero.  Hence,  if  a  represent  the  accelera- 
tion, 


*• 


76  APPLIED   MECHANICS. 

§  67.   Uniform  Motion In  this  case  the  acceleration  is 

zero,  and  the  velocity  is  constant ;  and  we  have  the  equation 


s  '=  vt. 


§68.   Uniformly  Varying  Motion.  —  In  this  case  the  ac- 
celeration is  constant :  hence  a  is  a  constant  in  the  equation 


and  we  obtain  by  one  integration 

.£*'£&$,, 

where  c  is  an  arbitrary  constant  :  to  determine  it  we  observe, 
that,  if  v0  represent  the  value  of  v  when  t  =  o,  we  shall  have 

V0  =   O  +  C 
.'.      C     =  V0 

,.  ,-£=*+«,   III      ^ 

and  by  another  integration 

.s  =    &P  4-  vj, 


where  s  is  the  space  passed  over  in  time  /  ;  the  arbitrary  con- 
stant vanishing,  because,  when  t  =  o,  s  is  also  zero. 

§  69.  Measure  of  Force.  —  It  has  already  been  seen,  that, 
when  a  body  is  either  at  rest  or  moving  uniformly  in  a  straight 
line,  there  are  either  no  forces  acting  upon  it,  or  else  the  forces 
acting  upon  it  are  balanced.  If,  on  the  other  hand,  the  motion 
of  the  body  is  rectilinear,  but  not  uniform,  the  only  unbalanced 
force  acting  is  in  the  direction  of  the  motion,  and  equal  in  mag- 
nitude to  the  momentum  imparted  in  a  unit  of  time  in  the  direc- 
tion of  the  motion,  or,  in  other  words,  to  the  limit  of  the  ratio 
of  the  momentum  imparted  in  a  short  time  (A*),  to  the  time,  as 
the  latter  approaches  zero. 


MECHANICAL    WORK.  —  UNIT  OF   WORK.  JJ 

Thus,  if  F  denote  the  force  acting  in  the  direction   of  the 
motion,  m  the  mass,  and  a  the  acceleration,  we  shall  have 

F=ma  =  m^=mm.  (0 

dt  dt2 

From  (i)  we  derive 

mdv  =  Fdt;  (2) 

and,  if  v0  be  the  velocity  of  the  moving  body  at  the  time  when 
/  =  /0,  and  vs  its  velocity  when  t  =  tlt  we  shall  have 


./,     «M 

or 


I   mdv  =  I 

*SZ>O  *J  tn 


ptl 

Vi  —  v0)  =  I   Fdt;  (3) 

«/*<> 


or,  in  words,  the  momentum  imparted  to  the  body  during  the 
time  /=(/,  —  /0)  by  the  force  F,  will  be  found  by  integrating 
the  quantity  Fdt  between  the  limits  tl  and  /0. 

§  70.  Mechanical  Work.  —  Whenever  a  force  is  applied  to 
a  moving  body,  the  force  is  either  use'd  in  overcoming  resist- 
ances (i.e.,  opposing  forces,  such  as  gravity  or  friction),  and 
leaving  the  body  free  to  continue  its  original  motion  undis- 
turbed, or  else  it  has  its  effect  in  altering  the  velocity  of  the 
body.  In  either  case,  the  work  done  by  the  force  is  the  prod- 
uct of  the  force,  by  the  space  passed  through  by  the  body  in 
the  direction  of  the  force. 

Unit  of  Work. — The  unit  of  work  is  that  work  which  is 
done  when  a  unit  of  force  acts  through  a  unit  of  distance  in 
the  same  direction  as  the  force ;  thus,  if  one  pound  and  one 
foot  are  our  units  of  force  and  length  respectively,  the  unit  of 
work  will  be  one  foot-pound. 

If  a  constant  force  act  upon  a  moving  body  in  the  direction 
of  its  motion  while  the  body  moves  through  the  space  s,  the 
work  done  by  the  force  is 

Fs; 


78  APPLIED  MECHANICS. 

and  this,  if  the  force  is  unresisted,  is  the  energy,  or  capacity  for 
performing  work,  which  is  imparted  to  the  body  upon  which  the 
force  acts  while  it  moves  through  the  space  s. 

Thus,  if  a  lo-pound  weight  fall  freely  through  a  height  of 
5  feet,  the  energy  imparted  to  it  by  the  force  of  gravity  during 
this  fall  is  10  X  5  =  50  foot-pounds,  and  it  would  be  necessary 
to  do  upon  it  50  foot-pounds  of  work  in  order  to  destroy  the 
velocity  acquired  by  it  during  its  fall.  If,  on  the  other  hand, 
the  force  is  a  variable,  the  amount  of  work  done  in  passing 
over  any  finite  space  in  its  own  direction  will  be  found  by  in- 
tegrating, between  the  proper  limits,  the  expression 


The  power  which  a  machine  exerts  is  the  work  which  it 
performs  in  a  unit  of  time. 

The  unit  of  power  commonly  employed  is  the  horse-power, 
which  in  English  units  is  equal  to  33000  foot-pounds  per 
minute,  or  550  foot-pounds  per  second. 

§71.  Energy.  —  The  energy  of  a  body  is  its  capacity  for 
performing  work. 

Kinetic  or  Actual  Energy  is  the  energy  which  a  body  pos- 
sesses in  virtue  of  its  velocity  ;  in  other  words,  it  is  the  work 
necessary  to  be  done  upon  the  body  in  order  to  destroy  its 
velocity.  This  is  equal  to  the  work  which  would  have  to  be 
done  to  bring  the  body  from  a  state  of  rest  to  the  velocity  with 
which  it  is  moving.  Assume  a  body  whose  mass  is  m>  and  sup- 
pose that  its  velocity  has  been  changed  from  VQ  to  vlf  Then  if 
F  be  the  force  acting  in  the  direction  of  the  motion,  we  shall 
have,  from  equation  (2),  §  69,  that 

Fvdt  —  mvdv;  (i) 

but 

vdt     =  ds 

.*.    Fds    =  mvdv.  (2) 


AT  WOOD'S  MACHINE.  79 

Hence,  by  integration, 

pvi  r 

I   mvdv  =    I  Fds 

*)VQ  *J 

.-.     \m(v?  -  V)  =  fFds;  (3) 

but  fFds  is  the  work  that  has  been  done  on  the  body  by  the 
force,  and  the  result  of  doing  this  work  has  been  to  increase 
its  velocity  from  VQ  to  vt.  It  follows,  that,  in  order  to  change 
the  velocity  from  v0  to  vlt  the  amount  of  work  necessary  to  per- 
form upon  the  body  is 

£F 

\m(p*  -  v<?)  =  J—  (vf  -  vj).  (4) 

g 

If  z/0  =  o,  this  expression  becomes 

}^,or^S  (5) 

which  is  the  expression  for  the  kinetic  energy  of  a  body  of  mass 
m  moving  with  a  velocity  vlt 

§  72.  Atwood's  Machine.  —  A  particular  case  of  uniformly 
accelerated  motion  is  to  be  found  in  Atwood's  machine,  in  which 
a  cord  is  passed  over  a  pulley,  and  is  loaded  with  unequal  weights 
on  the  two  sides.  Were  the  weights  equal,  there  would  be  no 
unbalanced  force  acting,  and  no  motion  would  ensue ;  but  when 
they  are  unequal,  we  obtain  as  a  result  a  uniformly  accelerated 
motion  (if  we  disregard  the  action  of  the  pulley),  because  we 
have  a  constant  force  equal  to  the  difference  of  the  two  weights 
acting  on  a  mass  whose  weight  is  the  sum  of  the  two  weights. 
Thus,  if  we  have  a  lo-pound  weight  on  one  side  and  a  5-pound 
weight  on  the  other,  the  unbalanced  force  acting  is 

F  =  10  —  5  =  5  Ibs. 


SO  APPLIED   MECHANICS. 


The  mass  moved  is  M  =  -  -t-5.  :  hence  the  resulting  ac- 

& 
celeration  is 


§  73.  Normal  and  Tangential  Components  of  the  Forces 
acting  on  a  Heavy  Particle.  —  If  a  body  be  in  motion,  either 
in  a  straight  or  in  a  curved  line,  and  if  at  a  certain  instant  all 
forces  cease  acting  on  it,  the  body  will  continue  to  move  at  a 
uniform  rate  in  a  straight  line  tangent  to  its  path  at  that  point 
where  the  body  was  situated  when  the  forces  ceased  acting. 

If  an  unresisted  force  be  applied  in  the  direction  of  the 
body's  motion,  the  motion  will  still  take  place  in  the  same 
straight  line  ;  but  the  velocity  will  vary  as  long  as  the  force 
acts,  and,  from  what  we  have  seen,  the  equation 

F=md-^L  (I) 

dt* 
will  hold. 

If  an  unresisted  force  act  in  a  direction  inclined  to  the 
body's  motion,  it  will  cause  the  body  to  change  its  speed,  and 
also  its  course,  and  hence  to  move  in  a  curved  line.  Indeed, 
if  a  force  acting  on  a  body  which  is  in  motion  be  resolved  into 
two  components,  one  of  which  is  tangent  to  its  xpath  and  the 
other  normal,  the  tangential  component  will  cause  the  body  to 
change  its  speed,  and  the  normal  component  will  cause  it  to 
change  the  direction  of  its  motion. 

The  measure  of  the  tangential  component  is,  as  we  have 
seen, 

77          d*s 
f  =  m  —  : 

dt2 

and  we  will  proceed  to  find  an  expression  for  the  normal  com- 
ponent  otherwise  known  as   the  Deviating  Force.      For  this 


CENTRIFUGAL   FORCE.  8 1 

purpose  we  may  substitute,  for  a  small  portion  of  the  curve,  a 
portion  of  the  circle  of  curvature ;  hence  we  will  proceed  to 
find  an  expression  for  the  centrifugal  force  of  a  body  which 
moves  uniformly  with  a  velocity  v  in  a  circle  whose  radius  is  r* 

CENTRIFUGAL    FORCE. 

Let  AC  (Fig.  47)  be  the  space  described  in  the  time  A£ 

Then  we  have  A  B 

AC  =  z;A/. 

The  motion  AC  may  be  approximately  consid- 
ered as  the  result  of  a  uniform  motion 


AB  —  v^t  nearly, 
and  a  uniformly  accelerated  motion 

BC  =  KM2  =  s, 
where  a  —  acceleration  due  to  centrifugal  force.     But 

(AB)2  =  BC  .  BD, 
or 

(e>A/)2  =  J^(A/)2(2r  +  s), 
where 

AO =  OC  =  r 

.•.     v2     =  \a(2r  -f  s)  approximately 

2V2 

/.     a       =  -  ' approximately. 

For  its  true  value,  pass  to  the  limit  where  s  =.  o. 

Hence  we  have,  for  the  acceleration  due  to  the  centrifugal 
force,  the  expression 

!? 

r 
Hence  the  centrifugal  force  is  equal  to 

r  gr  * 


82  APPLIED  MECHANICS. 


DEVIATING    FORCE. 

If  a  body  is  moving  in  a  curved  path,  whether  circular  or 
not,  and  the  unbalanced  force  acting  on  it  be  resolved  into  tan- 
gential and  normal  components,  the  tangential  component  will 
be,  as  has  already  been  seen, 

m — : 
<tt2 

and  the  normal  component  will  be 

tntf  =  m/dsV 
r          r\dt)  ' 

where  r  is  the  radius  of  curvature  of  the  path  at  the  point  in 
question. 

RESULTANT    FORCE. 

Hence  it  follows  that  the  entire  unbalanced  force  acting  on 
the  body  will  be 

or 


§  74.   Components  along  Three  Rectangular  Axes  of  the 
Velocities    of,   and    of    the    Forces    acting    on,   a    Moving 

Body. — If  we  resolve  the  velocity—  into  three  components 

along  OX,  OY,  and  OZ,  we  shall  have,  for  these  components 
respectively, 

**,      &,     and     ^; 
df      df  dt' 

this  being  evident  from  the  fact  that  dxt  dy,  and  dz  are  respec- 


COMPONENTS  OF   VELOCITIES  AND   FORCES.  83 

tively  the  projections  of  ds  on  the  axes  OX,  OY.t  and  OZ ;  and, 
from  the  differential  calculus,  we  have 


On  the  other  hand, 


dx       dy  -,     dz 

-dt-  i'  and  7, 


are  not  only  the  components  of  the  velocity  —  -  in  the  directions 

at 

OX,  OY,  and  OZ,  but  they  are  also  the  velocities  of  the  body 
in  these  directions  respectively. 

Now,  the  case  of  the  accelerations  is  different  ;  for,  while 


d2x       d2y  ,      d2z 

~      -     and     - 


are  the  accelerations  in  the  directions  OX,  OY,  and  (^respec- 
tively, they  are  not  the  components  of  the  acceleration 


dt2 
along  the  three  axes. 

That  they  are  the  former  is  evident  from  the  fact  that  -^, 

at 

JL  and   ~  are  the  velocities  in  the  directions  of  the  axes,  and 
dt  dt 

(i2v   d2/v   d2z 

-jj,  -7^,    —  are  their  differential  co-efficients,  and  hence  repre- 

sent the  accelerations  along  the  three  axes.     But  if  we  consider 
the  components  of  the  force  acting  on  the  body,  we  shall  have 


84  APPLIED   MECHANICS. 

for  its  components  along  OX,  OY,  and  OZ,  if  a,  J3,  and  y  are 
the  angles  made  by  F  with  the  axes  respectively, 

=  m&,       Fcosy  =  m—^ 


and  we  found  (§  73)  for  F,  the  value 

T     /  sJ^\  4. 

(») 


Hence,  equating  these  values  of  F,  and  simplifying,  we  shall 
have  the  equation 


Hence  it  is  plain  that  —  ^,  —  ^,  and  —  |  can  only  be  the  com- 
ponents of  the  actual  acceleration 


when  the  last  term  —  (  —  )  vanishes,  or  when  r  =  °o  ,  i.e.,  when 
r2\dtj 

the  motion  is  rectilinear. 

Moreover,  we  have  the  two  expressions  (i)  and  (2)  for  the 
force  acting  upon  a  moving  body. 

The  truth  of  the  proposition  just  proved  may  also  be  seen 
from  the  following  considerations  :  — 

If  a  parallelepiped  be  constructed  with  the  edges 

dx      dy      dz_ 

2?    5?    <tf 


CENTRIFUGAL   FORCE   OF  A   SOLID  BODY.  85 

the  diagonal  will  be  the  actual  velocity 

<& 
di 

and  will,  of  course,  coincide  in  direction  with  its  path. 

On  the  other  hand,  if  a  parallelepiped  be  constructed  with 

the  edges 

dzx      d2y      d2z 

dt2'      dt2'      dt2' 
its  diagonal  must  coincide  in  direction  with  the  force 


and  can  coincide  in  direction  with  the  path,  and  hence  with  the 

actual  acceleration 

d2s 
dt2' 

only  when  the  force  is  tangential  to  the  path,  and  hence  when 
the  motion  is  rectilinear. 

§75.   Centrifugal  Force  of  a  Solid  Body When  a  solid 

body  revolves  in  a  circle,  the  resultant  centrifugal  force  of  the 
entire  body  acts  in  the  direction  of  the  perpendicular  let  fall 
from  the  centre  of  gravity  of  the  body  on  the  axis  of  rotation, 
and  its  magnitude  is  the  same  as  if  its  entire  weight  were  con- 
centrated at  its  centre  of  gravity. 

PROOF.  —  Let  (Fig.  48)  the  angular  velocity  =  a,  and  the  total 
weight  =  W.  Assume  the  axis  of  rotation  perpendicular  to 
the  plane  of  the  paper  and  passing  through 
O  ;  assume,  as  axis  of  #,  the  perpendicular 
dropped  from  the  centre  of  gravity  upon 
the  axis  of  rotation.  The  co-ordinates  of 
the  centre  of  gravity  will  then  be  (ar0,  ^0), 
and  j/o  will  be  equal  to  zero. 

If,  now,  P  be  any  particle  of  weight  w, 
where  r  =  perpendicular  distance  from  P  on  axis  of  rotation, 


86  APPLIED   MECHANICS. 

and  x  =  OA,  y  =  AP,  we  shall  have  for  the  centrifugal  force 
of  the  particle  at  P 


but  if  we  resolve  this  into  two  components,  parallel  respectively 
to  OX  and  OY,  we  shall  have  for  these  components 

*  =  *wx     and      (™  a*  A*  =  a-wy, 
g         r        g  \g      Jr        g 

and,  for  the  resultant  for  the  entire  body  we  shall  have,  parallel 
to  OX, 

,  (i) 


g  g 

and 

Fy  =  —^wy   =  —  WyQ  =  o.  (2) 

cr  cr 

Hence  the  centrifugal  force  of  the  entire  body  is 

F=*Wx0-9  (3) 

g 

and  if  we  let  v0  =  a.x0  =  linear  velocity  of  the  centre  of  gravity, 
we  have 


which  is  the  same  as  though  the  entire  weight  of  the  body 
were  concentrated  at  its  centre  of  gravity. 


EXAMPLES. 

1.  A   lo-pound  weight  is  fastened  by  a  rope  5  feet  long  to  the 
centre,  around  which  it  revolves  at  the  rate  of  200  turns  per  minute ; 
find  the  pull  on  the  cord. 

2.  A  locomotive  weighing  50000  Ibs.,  whose  driving-wheels  weigh 
7000  Ibs.,  is  running  at  60  miles  per  hour,  the  diameter  of  the  drivers 


UNIFORMLY   VARYING   RECTILINEAR  MOTION.  87 

being  6  feet,  and  the  distance  from  the  centre  of  the  wheel  to  the  centre 
of  gravity  of  the  same  being  2  inches  (the  drivers  not  being  properly 
balanced)  ;  find  the  pressure  of  the  locomotive  on  the  track  (a)  when 
the  centre  of  gravity  is  directly  below  the  centre  of  the  wheel,  and  (^) 
when  it  is  directly  above. 

3.  Assume  the  same  conditions,  except  that  the  distance  between 
centre  of  the  wheel  and  its  centre  of  gravity  is  5  inches  instead  of  2. 

§76.  Uniformly  Varying  Rectilinear  Motion.  —  We  have 

already  found  for  this  case  (§  68)  the  equations 

d*s 

— -  =  a  =  a  constant, 

~    =  v  =  v0  +  at, 
at 

s      =  v0t  +  \at* ; 

and  we  may  write  for  the  force  acting,  which  is,  of  course,  coin- 
cident in  direction  with  the  motion, 

F  =  m —  =  ma  =  a  constant. 
dt2 

\ 

§  77.  Motion  of  a  Body  acted  on  by  the  Force  of  Gravity 
only.  —  A  useful  special  case  of  uniformly  varying  motion  is 
that  of  a  body  moving  under  the  action  of  gravity  only. 

The  downward  acceleration  due  to  gravity  is  represented  by 
g  feet  per  second,  the  value  of  g  varying  at  different  points  on 
the  surface  of  the  earth  according  to  the  following  law :  — 

g  =  gt(i  —  0.00284  cos  2\)fi  —  ^-\  feet  per  second, 

where 

gt  =  32.1695  feet, 

X   =  latitude  of  the  place, 

h   —  its  elevation  above  mean  sea-level  in  feet, 

R  =  20900000  feet. 


88  APPLIED   MECHANICS, 

If,  now,  we  represent  by  h  the  height  fallen  through  by  a 
descending  body  in  time  /,  we  shall  have  the  equations, 

v  =  v0  4-  gt, 


where  v0  is  the  initial  downward  velocity. 

If,  on  the  other  hand,  we  represent  by  v0  the  initial  upward 
velocity,  and  by  h  the  height  to  which  the  body  will  rise  in 
time  /  under  the  action  of  gravity  only,  we  must  write  the  equa- 
tions 

v  =  VQ  —  gt, 

h  =  v0t  - 


When  v0  =  o,  the  first  set  of  equations  gives 

v  —  gt, 


which  express  the  law  of  motion  of  a  body  starting  from  rest 
and  subject  to  the  action  of  gravity  only. 

Eliminate  t  between  these  equations,  and  we  shall  have 


or 


h  is  called  the  height  due  to  the  velocity  v,  and  represents  the 
height  through  which  a  falling  body  must  drop  to  acquire  the 
velocity  v  ;  and 


UNRESISTED  PROJECTILE.  89 

is  the  velocity  which  a  falling  body  will  acquire  in  falling 
through  the  height  h.  Thus,  if  a  body  fall  through  a  height  of 
50  feet,  it  will,  by  that  fall,  acquire  a  velocity  of  about 


=  ^3216.  66  =  56.7  feet  per  second. 

Again  :  if  a  body  has  a  velocity  of  40  feet  per  second,  we  shall 
have 

,       v2       1600  0  f    t 

h  =  —  —  --  =  24.8  feet  ; 
*g       64.3 

and  we  say  that  the  body  has  a  velocity  due  to  the  height  24.8 
feet,-  i.  e.,  a  velocity  which  it  would  acquire  by  falling  through  a 
height  of  24.8  feet. 


EXAMPLES. 

1 .  A  stone  is  dropped  down  a  precipice,  and  is  heard  to  strike  the 
bottom  in  4  seconds  after  it  started  :  how  high  is  the  precipice  ? 

2.  How  long  will  a  stone,  dropped  down  a  precipice  500  feet  high, 
take  to  reach  the  bottom? 

3.  What  will  be  its  velocity  just  before  striking  the  ground? 

4.  A  body  is  thrown  vertically  upwards  with  a  velocity  of  100  feet 
per  second  ;  to  what  height  will  it  rise  ? 

5.  A  body  is  thrown  vertically  upwards,  and  rises  to  a  height  of  50 
feet.     With  what  velocity  was  it  thrown,  and  how  long  was  it  in  its 
ascent  ? 

6.  What  will  be  its  velocity  in  its  ascent  at  a  point  15  feet  above 
the    point  from  which   it   started,  and  what   at   the  same  point  in  its 
descent  ? 

§  78.  Unresisted  Projectile.  —  In  the  case  of  an  unresisted 
projectile,  we  have  a  body  on  which  is   impressed  a  uniform 


9o 


APPLIED   MECHANICS. 


motion  in  a  certain  direction  (the  direction  of  its  initial  motion), 
and  which  is  acted  on  by  the  force  of  gravity  only. 

Y  Let    OPC    be 

,A  ^  the  path  (Fig.  49), 

OA  the  initial  di- 
rection, and  v0  the 
initial  velocity,  and 
the  angle  AOX — 
c  0. 

Then  we  shall 
have,  for  the  hori- 
zontal and  vertical 
components  of  the  unbalanced  force  acting,  when  the  projectile 
is  at  P  (co-ordinates  x  and  y), 

sfZ  *£  /J2  +  1 

m  — -  ==  o  along  OX,  and  m-~  =  —mg  =  —W  along  O  Y. 
Hence 

£=°'    (i)     ^=-g-  .:•:  •„    (a) 

Integrating,  and  observing,  that,  when  t  =  o,  the  horizontal 
and  the  vertical  velocities  were  respectively  ^Ocos  6  and  ^Osin  0, 
we  have 


FIG.  49. 


dt 


=  v0  cos 


-j    =  ^0  sn      - 
dt 


(3) 

(4) 


These  equations  could  be  derived  directly  by  observing  that 
the  horizontal  component  of  the  initial  velocity  is  v0  cos  0,  and 
that  this  remains  constant,  as  there  is  no  unbalanced  force  act- 
ing in  this  direction,  also  that  v0  sin  0  is  the  initial  vertical 
velocity ;' and,  since  the  body  is  acted  on  by  gravity  only,  this 
velocity  will  in  time  t  be  decreased  by  gt. 


UNRESISTED   PROJECTILE.  9! 

Integrating  equations  (3)  and  (4),  and   observing  that  for 
t  —  o,  x  and  y  are  both  zero,  we  obtain 

x  =  v0  cos  0./,  (5) 

y  =  VQ  sin  6.t  -  %gt\  (6) 
Eliminate  /,  and  we  have 

j  =  *tan0  --  ^—  (7) 

2VQ2  COS2  0 

as  the  equation  of  the  path,  which  is  consequently  a  parabola. 

Equations  (i),  (2),  (3),  (4),  (5),  (6),  and  (7)  enable  us  to  solve 
any  problem  with  reference  to  an  unresisted  projectile. 

Equation  (7)  may  be  written 


/    _ 
V  ' 


fro2  sin2  B\  g  fro2  sin  0  cos  0 


2g        /  2V<?  COS2  0  \  g  / 

which  gives  for  the  co-ordinates  of  the  vertex 

_  v02  sin2  0  _  fr02  sin  6  cos  0 

•    2g  g 

EXAMPLES. 

i.  An  unresisted  projectile  starts  with  a  velocity  of  100  feet  per 
second  at  an  upward  angle  of  30°  to  the  horizon  ;  what  will  be  its  velocity 
when  it  has  reached  a  point  situated  at  a  horizontal  distance  of  1000  feet 
from  its  starting-point,  and  how  long  will  be  required  for  it  to  reach 
that  point? 

Solution. 

v0  =  100,        0  =  30°,        fr0  cos  6  =  86.6,        v0  sin  0  =  50, 

g  =  32.16. 
Equation  (5)  gives  us 

1000  =  86.6  t 

/.    /=  I00°  =  11.55  seconds. 
86.6 


APPLIED   MECHANICS. 


sm 


1  -  gt  =  5°  -  37i-5  =  -32I-5> 
+  (32I-5)2  =  ^75°°  +  103362  =  333- 


Hence  the  point  in  question  will  be  reached  in   n|-  seconds  after  start- 
ing, and  the  velocity  will  then  be  333  feet  per  second. 

*2.  An  unresisted  projectile  is  thrown  upwards  from  the  surface  of 
the  earth  at  angle  of  39°  to  the  horizontal :  find  the  time  when  it  will 
reach  the  earth,  and  the  velocity  it  will  have  acquired  when  it  reaches 
the  earth,  the  velocity  of  throwing  being  30  feet  per  second. 

3.  A  lo-pound  weight  is  dropped  from  the  window  of  a  car  when 
travelling  over  a  bridge  at  a  speed  of  25  miles  an  hour.     How  long  will 
it  take  to  reach  the  ground  100  feet  below  the  window,  and  what  will  be 
the  kinetic  energy  when  it  reaches  the  ground  ? 

4.  With  what  horizontal  velocity,  and  in  what  direction,  must  it  be 
thrown,  in  order  that  it  may  strike  the  ground  50  feet  forward  of  the 
point  of  starting? 

5.  Suppose  the  same  lo-pound  weight  to  be  thrown  vertically  up- 
wards from  the  car  window  with  a  velocity  of  TOO  feet  a  minute,  how 
long  will  it  take  to  reach  the  ground,  and  at  what  point  will  it  strike  the 
ground  ? 

§  79.  Motion  of  a  Body  on  an  Inclined  Plane  without 

Friction.  —  If  a  body  move  on 
an  inclined  plane  along  the  line 
of  steepest  descent,  subject  to 
the  action  of  gravity  only,  and 
if  we  resolve  the  force  acting 
on  it  (i.e.,  its  weight)  into  two 
components,  along  and  perpen- 
dicular to  the  plane  respec- 
tively, the  latter  component 
will  be  entirely  balanced  by 
the  resistance  of  the  plane, 

and  the  former  will   be  the  only  unbalanced   force  acting  on 

the  body. 


MOTION  OF  A   BODY  ON  AN  INCLINED   PLANE.  93 

Suppose  a  body  whose  weight  is  represented  (Fig.  50)  by 
HF  =  W  to  move  along  the  inclined  path  AB  under  the  action 
of  gravity  only.  Let  6  be  the  inclination  of  AB  to  the  horizon. 
Resolve  J^into  two  components, 

HD  =  JVs'm  <9,     and     HE  =  W  cos  0, 

respectively  parallel  and  perpendicular  to  the  plane.  The 
former  is  the  only  unbalanced  force  acting  on  the  body,  and 
will  cause  it  to  move  down  the  plane  with  a  uniformly  accel- 
erated motion  ;  the  acceleration  being  , 


=  g  sin  0.  (i) 


(f) 


If  the  body  is  either  at  rest  or  moving  downwards  at  the 
beginning,  it  will  move  downwards ;  whereas,  if  it  is  first  mov- 
ing upwards,  it  will  gradually  lose  velocity,  and  move  upwards 
more  slowly,  until  ultimately  its  upward  velocity  will  be  de- 
stroyed, and  it  will  begin  moving  downwards. 

The  equations  for  uniformly  varying  motion  are  entirely 
applicable  to  these  cases.  Thus,  suppose  that  the  body  has  an 
initial  downward  velocity  t/0,  this  .velocity  will,  at  the  end  of  the 
time  t,  become 

z>  =  ^  =  v0  +  tesinfl)/  (2) 

at 

/.     s  =  vQt  4-  \g  sin  0  .  t*,  (3) 

and,  for  the  unbalanced  force  acting,  we  have 

j?=m^l  =  J^sintf)  =  WsinO.  (4) 

dt*        ? 


94  APPLIED   MECHANICS. 

If,  on  the  other  hand,  the  body's  initial  velocity  is  upward, 
and  we  denote  this  upward  velocity  by  v0,  we  shall  have  the 
equations 

v  =  f  =  v0  -  (gsmO)t        V          (5) 
at 

sin  6  .  /2  (6) 

(7) 

Again,  if  the  initial  velocity  is  zero,  equations  (2)  and  (3) 
become 


s  =  ksin<9  .  /2.  (9) 

From  these  we  obtain,  for  this  case, 

/  =  J    2S     ;  (10) 

and,  substituting  this  value  of  t  in  (8),  we  have 


v  —  \*g(s  sin^),  (n) 

or,  if  we  let  s  sin  6  =  h  •=.  the  vertical  distance  through  which 
the  body  has  fallen,  we  have 

v  =  ^2gh.  (12) 


Hence,  When  a  body,  starting  from  rest,  falls,  under  the 
action  of  gravity  only,  through  a  height  h,  the  velocity  acquired 
is  V/2gh,  whether  the  path  be  vertical  or  inclined. 


EXAMPLES. 


i.  A  body  moves  from  the  top  to  the  bottom  of  a  plane  inclined 
to  the  horizon  at  30°,  under  the  action  of  gravity  only  :  find  the  time 
required  for  the  descent,  and  the  velocity  at  the  foot  of  the  plane. 


MOTION  ALONG   A    CURVED   LINE. 


95 


FIG.  51. 


2.  In  the  right-angled  triangle  shown  in  the  figure  (Fig.  51),  given 
AB  =  10  feet,   angle  BAG  =  30°:    find  the  time  a 

body  would  require,  if  acted  on  by  gravity  only,  to  fall 
from  rest  through  each  of  the  sides  respectively,  AB 
being  vertical. 

3.  Given  inclination  of  plane  to  the  horizon  =  6, 
length  of  plane  —  //  compare  the  time  of  falling  down 
the  plane  with  the  time  of  falling  down  the  vertical. 

4.  A  loo-pound  weight  rests,  without  friction,  on  the 
plane  of  example  3.     What  horizontal  force  is  required 
to  keep  it  from  sliding  down  the  plane. 

5.  Suppose  5  pounds  horizontal  force  to  be  applied 

(a)  so  as  to  oppose  the  descent,  (£)  so  as  to  aid  the  descent  :  find  in 
each  case  how  long  it  will  take  the  weight  to  descend  from  the  top  to 
the  bottom  plane. 

§  80.  Motion  along  a  Curved  Line  under  the  Action  of 
Gravity  only.  —  We  shall  consider  two  questions  in  this 
regard  :  (a)  the  velocity  at  any  point  of  the  curve  (b)  the  time 
of  descent  through  any  part  of  the  curve. 

(a)  Velocity  at  any  point.     Let  us  suppose  the  body  to  have 

started  from  rest  at  A,  and  to  have 
reached  the  point  P  in  time  /, 
where  AB  —  x  (Fig.  52).  Then, 
since  the  curved  line  AP  may  be 
considered  as  the  limit  of  a  broken 
line  running  from  A  to  Py  and  as 
it  has  already  been  seen  that  the 
velocity  acquired  by  falling  through 
"c  a  certain  height  depends  only  upon 
the  height,  and  not  upon  the  incli- 
nation of  the  path,  we  shall  have  for  a  curved  line  also 


V  = 


where  v  is  the  velocity  at  P. 


FIG.  52. 


96 


APPLIED   MECHANICS. 


(b)  Time  doivn  a  curve.  Referring  to  the  same  figure,  let  t 
denote  the  time  required  to  go  from  A  to  P,  and  A^  the  time  to 
go  from  P  to  P't  where  PP'  —  AJ,  and  BB'  —  kx  ;  then,  as  we 
have  seen  that  the  velocity  at  P  is  ^2gx,  we  shall  have  approx- 
imately for  the  space  passed  over  in  time  A/,  the  equation 


or,  passing  to  the  limit, 


This  equation  gives 


ds 


f  ^ 

(0 


or 


=  r==  r 

J^2gX  J 


where,   of  course,   the   proper  limits   of   integration   must   be 
used. 

If  t  denote  the  time  from  A  to  P,  we  have 


2gX 


EXAMPLE. 

A  body  acted  on  by  gravity  only  is  constrained  to 
move  in  the  arc  of  a  circle  from  A  to  C  (Fig.  53),  radius 
10  feet.  Find  the  time  of  describing  the  arc  (quadrant) 
and  the  velocity  acquired  by  the  body  when  it  reaches 


FIG.  53. 


SIMPLE   CIRCULAR  PENDULUM. 


97 


§8 1.  Simple  Circular  Pendulum.  —  To  find  the  time  occu- 
pied in  a  vibration  of  a  simple  circu- 
lar  pendulum,  we  take  D  (Fig.  54)  as 
origin,  and  DC  as  axis  of  x,  and  the 
axis  of  jj/at  right  angles  to  DC.  Let 
AC  —  /and  BD  —  h,  we  shall  have 
for  the  time  of  a  single  oscillation 
from  A  to  E 


-f. 


ds 


D 
FIG.  54. 


Now,  from  the  equation  of  the  circle  AFDE, 

y2  =  2lx  —  x2, 


we  have 


dy_  =  I  -  x 
dx          y 

dl-  =  l-  =  / 

~dx~  y~  v/2  lx  -  x"- 


Idx 


or 


2/ 


This  can  only  be  integrated  approximately. 
Expanding  (  I  —  —  j        we  obtain 

(i  -  -Y *  =  J  +  ~7  +  -y'  +  etc., 

\      2/y  4/     32  /2 

//  rk  /      x      -\  x< 

•'•  '  =  V;./  (I  +  r/  +  ^? 


dx 


98  APPLIED   MECHANICS. 

The  greatest  value  of  x  is  //;  and  if  h  is  so  small  that  we  may 

omit  — ,  we  shall  have  as  our  approximate  result 
4/ 


//     Ck  dx  1 1  (  -  '  2X }  h 

\~o\      J,  a  =  \Hversm      ~\     =  ^/  -.       (i) 

"  8J       \hx  —  x2 


,    ••"       =  v  - \ versin     =r\    =  *V  -• 

!hx  -  x2       V^(  /i  }0          \  g 


If,  however,  the  value  of  h  as  compared  with  /  is  too  large 


to  render  it  sufficiently  accurate  to  omit  —  ,  but  so  small  that 

4/ 

we  can  safely  omit  the  higher  powers  of  -,  we  shall  have 


/(  -1  2*    ,     i    C 

=  vMversm     -r-+-—,1 

V^l  h        441 


\//ix-x2 

°*  +  ±[* 
h        4/L2 


or 

/  = 

a  nearer  approximatioa 
The  formula 


/  = 


is  the  most  used,  and  is  more  nearly  correct,  the  smaller  the 
value  of  h. 

EXAMPLES. 

i  .  Find  the  length  of  the  simple  circular  pendulum  which  is  to  beat 
seconds  at  a  place  where  g  = 


Solution. 


SIMPLE    CYCLOIDAL   PENDULUM. 


99 


2.  What  is  the  time  of  vibration  of  a  simple  circular  pendulum  5 
feet  long? 

§82.   Simple  Cycloidal  Pendulum. — The  equation  of  the 
cycloid  is 

y  = 


.     *y  =  J2a  -  x 

dx       V       x 

.:     *  =  (™\\ 
dx       \x) 

Hence  we  shall  have,  for  the  time  of  a  single  oscillation, 


or 

/=  2(  -  )  <;versm      — -j.  = 
W  (  h    . 

This  expression  is-  independent  of  /^,  so  that  the  time  of  vibra- 
tion is  the  same  whether  the  arc  be  large  or  small. 

A  body  can  be  made  to  vibrate  in  a  cycloidal  arc  by  suspend- 
ing it  by  a  flexible  string  between  two  cycloidal  cheeks.  This 
is  shown  from  the  fact  that 
the  evolute  of  the  cycloid  is 
another  cycloid  (Fig.  55). 

To  prove  this,  we  have, 
from  the  equation  of  the 
cycloid, 

><' 


dv  _     /2a  —  x     ds  _     1 2a 
~dx~\  ~~^~'    ~dx  =  \^ 


—  a 


&  ^ 

dx7-       xl^2a  _  x 


100  APPLIED  MECHANICS. 


Hence  the  radius  of  curvature  is 


dx2 

and  since  we  have  for  the  evolute  the  relation 

ds'  =  dp, 
where  ds'  is  the  elementary  arc  of  the  evolute, 

f*X  =  20. 

.:    /=    /    dp; 

J  x  =  x 

and,  observing  that  when  x  =  2a    p  =  o,    we  have 


/.         =  22020  —  x. 
If  xl  is  the  abscissa  of  the  point  of  the  evolute, 


.'.       /     =    2(2dYXl    —    20,  ; 

and,  transforming  co-ordinates  to  ^  by  putting  ^r2  +  2^  for 
we  obtain 


which  is  the  equation  of  another  cycloid  just  like  the  first 

The  motion  along  a  vertical  cycloid  may  also  be  obtained  by 
letting  a  body  move  along  a  groove  in  the  form  of  a  cycloid 
acted  on  by  gravity  alone  ;  and  in  this  case  the  time  of  descent 
of  the  body  to  the  lowest  point  is  precisely  the  same  at  what- 
ever point  of  the  curve  the  body  is  placed. 

§  83.  Effect  of  Grade  on  the  Tractive  Force  of  a  Rail- 
way Train.  —  As  a  useful  particular  case  of  motion  on  an 
inclined  plane,  we  have  the  case  of  a  railroad  train  moving  up 
or  down  a  grade.  It  is  necessary  that  a  certain  tractive  force 


EFFECT  OF  GRADE   ON  TRACTIVE   FORCE.  IOI 

be  exerted  in  order  to  overcome  the  resistances,  and  keep 
the  train  moving  at  a  uniform  rate  along  a  level  track.  If, 
on  the  other  hand,  the  track  is  not  on  a  level,  and  if  we 
resolve  the  weight  of  the  train  into  components  at  right  angles 
to  and  along  the  plane  of  the  track,  we  shall  have  in  the  latter 
component  a  force  which  must  be  added  to  the  tractive  force 
above  referred  to  when  we  wish  to  know  the  tractive  force  re- 
quired to  carry  it  up  grade,  and  must  be  subtracted  when  we 
wish  to  know  the  tractive  force  required  to  carry  it  down  grade. 
The  result  of  this  subtraction  may  give,  if  the  grade  is  suffi- 
ciently steep  and  the  speed  sufficiently  slow,  a  negative  quan- 
tity ;  and  in  that  case  we  must  apply  the  brakes,  instead  of 
using  steam,  unless  we  wish  the  speed  of  the  train  to  increase. 

EXAMPLES. 

i.  A  railroad  train  weighing  60000  Ibs.,  and  running  at  50  miles  per 
hour,  requires  a  tractive  force  of  618  Ibs.  on  a  level ;  what  is  the  tractive 
force  necessary  when  it  is  to  ascend  a  grade  of  50  feet  per  mile?  What 
when  it  is  to  descend?  Also  what  is  the  amount  of  work  per  minute 
in  each  case  ? 

Solution. 

The  resolution  of  the  weight  will  give  (Fig.  50,  §  79),  for  the  com- 
ponent along  the  plane, 

(60000) -^  =  568.2  nearly. 
Hence 

Tractive  force  for  a  level  =  618.0, 
Tractive  force  for  ascent  =  1186.2, 
Tractive  force  for  descent  =  49.8. 

To  ascertain  the  work  done  per  minute  in  each  case,  we  have  — 
(a)  For  a  level  track,  6l8  x  5^x  528°  =  2719200  foot-lbs. 

(t)  Up  grade,       2719200  +  6ooo°  X6o5°  x  5°  =  5219200  foot-lbs. 
(c)  Down  grade,  2719200  -  6ooo°  ^  x  5°  =     219200  foot-lbs. 


IO2 


APPLIED   MECHANICS. 


2.  Suppose  the  tractive  force  required  for  each  2000  Ibs.  of  weight 
of  train  to  be,  on  a  level  track,  for  velocities  of — 

5.0  miles  per  hour,     10.0     20.0     30.0     40.0     50.0     60 

6.1  Ibs.,  6.6       8.3     ii. 2     15.3     20. 6     27; 
find  the  tractive  force  required  to  carry  the  train  of  example  i  — 

(a)  Up  an  incline  of  50  feet  per  mile  at  30  miles  per  hour. 
(&)   Down  an  incline  of  50  feet  per  mile  at  30  miles  per  hour. 
(<:)  Down  an  incline  of  10  feet  per  mile  at  20  miles  per  hour. 
(d)  What  must  be  the  incline  down  which  the  train  must  run  to 
require  no  tractive  force  at  40  miles  per  hour? 

3.  If  in  the  first  example  the  tractive  force  remains  618  Ibs.  while 
the  train  is  going  down  grade,  what  will  be  its  velocity  at  the  end  of  one 
minute,  the  grade  being  10  feet  per  mile? 

§  84.    Harmonic    Motion.  —  If  we  imagine  a  body  to  be 
moving  in  a  circle  at  a  uniform  rate  (Fig.  56),  and  a  second 

body  to  oscillate  back  and  forth  in 

the    diameter    AB,    both     starting 

from  B,    and 

if    when    the 

first    body  is 
'x    at  C  the  other 

is  directly  un- 
der  it    at   G> 

etc.,   then   is 

the       second 

body  said   to 
move  in  harmonic  motion. 

A  practical  case  of  this  kind  of  mo- 
tion is  the  motion  of  a  slotted  cross-head 
of  an  engine,  as  shown  in  the  figure 
(Fig.  57) ;  the  crank  moving  at  a  uni- 
form rate.  In  the  case  of  the  ordinary 
crank,  and  connecting-rod  connecting 
the  drive-wheel  shaft  of  a  stationary  engine  with  the  piston-rod, 


FIG.  56. 


HARMONIC  MOTION.  1  03 

_  ________  _  ^   _____ 

we  have  in  the  motion  of  the  piston  only  an  approximation  to 
harmonic  motion.  We  will  proceed  to  determine  the  law  of  the 
force  acting  upon,  and  the  velocity  of,  a  body  which  is  con- 
strained to  move  in  harmonic  motion.  Let  the  body  itself  and 
the  corresponding  revolving  body  be  supposed  to  start  from 
B  (Fig.  56),  the  latter  revolving  in  left-handed  rotation  with  an 
angular  velocity  a,  and  let  the  time  taken  by  the  former  in 
reaching  G  be  t:  then  will  the  angle  BOC  =  at;  and  we  shall 
have,  if  s  denote  the  space  passed  over  by  the  body  that  moves 
with  harmonic  motion, 

s  =  BG  =  OB  -  OCcosat, 
or,  if 

r  =  OB  =  OC, 

s  =  r  —  r  cos  at,  >,*{..  (i) 

the  velocity  at  the  end  of  the  time  /  will  be 

V  =  —  =  arsmat,  (2) 

and  the  acceleration  at  the  end  of  time  t  will  be 


Hence  the  force  acting  upon  the  body  at  that  instant,  in  the 
direction  of  its  motion,  is 

//2c 
F  =  m—2  =  ma.2  r  cos  at  =  ma2(OG).  (4) 

The  force,  therefore,  varies  directly  as  the  distance  of  the.  body 
from  the  centre  of  its  path.     It  is  zero  when  the  body  is  at  the 


104  APPLIED  MECHANICS. 

centre  of  its  path,  and  greatest  when  it  is  at  the  ends  of  its 
travel,  as  its  value  is  then 

W 
ma2r  =  —a2r; 

g 

this  being  the  same  in  amount  as  the  centrifugal  force  of  the 
revolving  body,  provided  this  latter  have  the  same  weight  as  the 
oscillating  body.  On  the  other  hand,  the  velocity  is  greatest 

when  at  =  -  (i.e.,  at  mid-stroke) ;  and  its  value  is  then 

V  =  ar, 

this  being  also  the  velocity  of  the  crank-pin  at  mid-stroke. 

EXAMPLE. 

Given* that  the  reciprocating  parts  of  an  engine  weigh  10000  Ibs., 
the  length  of  crank  being  i  foot,  the  crank  making  60  revolutions  per 
minute  ;  find  the  force  required  to  make  the  cross-head  follow  the  crank, 
(i)  when  the  crank  stands  at  30°  to  the  line  of  dead  points^ (2)  when 
at  60°,  (3)  when  at  the  dead  point. 

§85.  Work  under  Oblique  Force.  —  If  the  force  act  in 
any  other  direction  than  that  of  the  motion,  we  must  resolve  it 
into  two  components,  the  component  in  the  direction  of  the 
motion  being  the  only  one  that  does  work.  Thus  if  the  force 
F  is  variable,  and  0  equals  the  angle  it  makes  with  the  direction 
of  the  motion,  we  shall  have  as  our  expression  for  the  work 

done 

fF  cos  Qds. 

Thus  if  a  constant  force  of  100  Ibs.  act  upon  a  body  in  a  direc- 
tion making  an  angle  of  30°  with  the  line  of  motion,  then  will 
the  work  done  by  the  force  during  the  time  in  which  it  moves 
through  a  distance  of  10  feet  be 

(100)  (0.86603)  (10)  =  866foot-lbs. 


ROTATION  OF  RIGID   BODIES.  105 

§86.  Rotation  of  Rigid  Bodies.  —  Suppose  a  rigid  body 
(Fig.  58)  to  revolve  about  an  axis  perpendicular  to  the  plane  of 
the  paper,  and  passing  through  O  ; 
imagine  a  particle  whose  weight  is 
w  to  be  situated  at  a  perpendicular 
distance  OA  =.  r  from  the  axis  of 
rotation,  and  let  the  angular  velocity 
be  a  :  let  it  now  be  required  to  find 
the  moment  of  the  force  or  forces 
required  to  impart  this  motion  in  a 
unit  of  time  ;  for  we  know,  that,  if 

the  axis  of  rotation  pass  through  the  centre  of  gravity  of  the 
body,  the  motion  can  be.  imparted  only  by  a  statical  couple; 
whereas  if  it  do  not  pass  through  the  centre  of  gravity,  the 
motion  can  be  imparted  by  a  single  force. 

We  shall  have,  for  the  particle  situated  at  A, 

Weight  =  w. 

Angular  velocity  =  a. 

Linear  velocity  =  ar. 

Force  required  to  impart  this  velocity  in  a  unit  of  time  to 

ngj 

this  particle  =  —  ar. 
g 

Moment  of  this  force  about  the  axis  =  —  ar2. 

S 

Hence  the  moment  of  the  force  or  forces  required  to  impart 
to  the  entire  body  in  a  unit  of  time  a  rotation  about  the  axis 
through  0,  with  an  angular  velocity  a,  is 


g       g          g 

where  /  is  used  as  a  symbol  to  denote  the  limit  of  'Stvr2,  and  is 
called  the  Moment  of  Inertia  of  the  body  about  the  axis  through  O. 


106  APPLIED   MECHANICS. 

§87.   Angular  Momentum.  —  This  quantity,—,  which  ex- 

g 

presses  the  moment  of  the  force  or  forces  required  to  impart  to 
the  body  in  a  unit  of  time  the  angular  velocity  a  about  the  axis 
in  question  is  also  called  the  Angular  Momentum  of  the  body 
when  rotating  with  the  angular  velocity  a  about  the  given 
axis. 

§  88.  Actual  Energy  of  a  Rotating  Body.  —  If  it  be  required 
to  find  the  actual  energy  of  the  body  when  rotating  with  the 
angular  velocity  a,  we  have,  for  the  actual  energy  of  the  particle 
at  A, 

W  (ar)2         a2 

—  -  —  ~  =  —  wr2, 

g        2  2g       . 

and  for  that  of  the  entire  body 

a2/ 


This  is  the  amount  of  mechanical  work  which  would  have  to  be 
done  to  bring  the  body  from  a  state  of  rest  to  the  velocity  a,  or 
the  total  amount  of  work  which  the  body  could  do  in  virtue 
of  its  velocity  against  any  resistance  tending  to  stop  its 
rotation. 

§  89.  Moment  of  Inertia.  —  The  term  "moment  of  inertia" 
originated  in  a  wrong  conception  of  the  properties  of  matter. 
The  term  has,  however,  been  retained  as  a  very  convenient  one, 
although  the  conceptions  under  which  it  originated  have  long 
ago  vanished.  The  meaning  of  the  term  as  at  present  used,  in 
relation  to  a  solid  body,  is  as  follows  :  — 

The  moment  of  inertia  of  a  body  about  a  given  axis  is  the 
limit  of  the  sum  of  tJie  products  of  the  weight  of  each  of  the  ele- 
mentary particles  that  make  up  the  body,  by  the  squares  of  their 
distances  from  the  given  axis. 

Thus,  if  w,,  w2,  wy  etc.,  are  the  weights  of  the  particles 
which  are  situated  at  distances  rlt  r2,  ry  etc.,  respectively  from 


MOMENT  OF  INERTIA    OF  A   PLANE  SURFACE.  IO/ 

the  axis,  the  moment  of  inertia  of  the  body  about  the  given 

axis  is 

/  ==  limit  of 


§90.  Radius  of  Gyration.  —  The  radius  of  gyration  of  a 
body  with  respect  to  an  axis  is  the  perpendicular  distance  from 
the  axis  to  that  point  at  which,  if  the  whole  mass  of  the  body 
were  concentrated,  the  angular  momentum,  and  hence  the  mo- 
ment of  inertia,  of  the  body,  would  remain  the  same  as  they  are 
in  the  body  itself. 

If  p  is  the  radius  of  gyration,  the  moment  of  inertia  would 
be,  when  the  mass  is  concentrated, 

P2io// 
hence  we  must  have 

p22o>  =  ^wr2  =  /, 
whence 

2        ^wr2  =    _/ 
''    2w    :=  W9 

where  W  =  entire  weight  of  the  body. 

§91.  Moment  of  Inertia  of  a  Plane  Surface.  —  The  term 
"moment  of  inertia,"  when  applied  to  a  plane  figure,  must,  of 
course,  be  defined  a  little  differently,  as  a  plane  surface  has  no 
weight  ;  but,  inasmuch  as  the  quantity  to  which  that  name  is 
given  is  necessary  for  the  solution  of  a  great  many  questions. 

The  moment  of  inertia  of  a  plane  surface  about  an  axis,  either 
in  or  not  in  the  plane  y  is  the  limit  of  .the  sum  of  the  products  of 
the  elementary  areas  into  which  the  surface  may  be  conceived  to 
be  divided,  by  the  squares  of  their  distances  from  the  axis  in 
question. 

In  a  similar  way,  for  the  radius  of  gyration  p  of  a  plane 
figure  'whose  area  is  A,  we  have 


108  APPLIED   MECHANICS. 

From  this  definition  it  will  be  evident,  that,  if  the  surface  be 
referred  to  a  pair  of  axes  in  its  own  plane,  the  moment  of  iner- 
tia of  the  surface  about  O  Y  will  be 


(i) 

and  the  moment  of  inertia  of  the  surface  about  OX  will  be 

J^fffdxdy.  (2) 

The  moment  of  inertia  of  the  surface  about  an  axis  passing 
through  the  origin,  and  perpendicular  to  the  plane  XO  Y,  will  be 

fS^dxdy,  (3) 

where  r=  distance  from  O  to  the  point  (x,y)\  hence  r*  =  x*  + 
^2,  and  the  moment  of  inertia  becomes 

ff(x2  +  f)dxdy  =  fftfdxdy  4-  fffdxdy  =  /  +  /.      (4) 

This  is  called  the  "polar  moment  of  inertia."  If  polar  co-ordi- 
nates be  used,  this  last  becomes 

ffp(pdpdB)  =  ffp^dpdd.  (5) 

All  these  quantities  are  quantities  that  will  arise  in  the  discus- 
sion of  stresses,  and  the  letters  /and  /are  very  commonly  used 
to  denote  respectively 

ffx*dxdy        and        fffdxdy. 

Another  quantity  that  occurs  also,  and  which  will  be  repre- 
sented by  K,  is 

Sfxydxdy;  (6) 

and  this  is  called  the  moment  of  deviation. 


EXAMPLES   OF  MOMENTS   OF  INERTIA. 


109 


EXAMPLES. 

The  following   examples   will   illustrate   the   mode   of  finding   the 
moment  of   inertia  :  —  x 

i.  Find  the  moment  of  inertia  of  the  rectangle 
ABCD  about  OY  (Fig.  59). 


Solution. 


FIG.  59. 


2.  Find  the  moment  of  inertia  of  the  entire  circle  (radius  r)  about 
the  diameter  OY  (Fig.  60). 


FIG.  60. 


Solution. 


/V      /»V,.3_J|:a  (* 

/  =  J    J     _  x*dxdy  =  2j 
r*  f  V 

At/ 


-  x*dx 


=  21 x\  r  — 

4 


64  ' 


3.  Find  the  moment  of  inertia  of  the  circular  ring  (outside  radius  r, 
inside  radius  r^)  about  OY  (Fig.  61). 


Solution. 


"44 


4.  Find  the  moment  of  inertia  of  an  ellipse 
(semi-axes  a  and  b}  about  the  minor  axis  OY. 


64 


FIG.  61. 


I  IO 


APPLIED   MECHANICS. 


Solution. 


x2       y2 
Equation  of  ellipse  is  —  +  -^  =  i 


, 
(*(i      f*a 

y  =  J  J  , 

~n  ' 


-  x'dx  = 


',  8 


On  the  other  hand,  Ix  — 


§  92.  Moments  of  Inertia  of  Plane  Figures  about  Parallel 
Axes. 

PROPOSITION.  —  The  moment  of  inertia  of  a  plane  figure 
about  an  axis  not  passing  through  its  centre  of  gravity  is  equal 
to  its  moment  of  inertia  about  a  parallel  axis  passing  through  its 
centre  of  gravity  increased  by  the  product  obtained  by  multiply- 
ing the  area  by  the  square  of  the  distance  between  the  two  axes. 

PROOF.  —  Let  A  B  CD 
(Fig.  62)  be  the  surface ;  let 
OY  be  the  axis  not  passing 
through  the  centre  of  grav- 
ity ;  let  P  be  an  elementary 


YI        *  *"*  '         Tl         area   A;rAj/,    whose   co-ordi- 

nates are  OR  =  x  and  RP 
=  y  ;  and  let  OOl  —  a  =  a 
constant    =    distance     be- 
tween the  axes. 
Let  O^R  =  xl  =  abscissa  of  P  with  reference  to  the  axis 
passing  through  the  centre  of  gravity, 


x  =  a  -f 
x2  —  x?  + 


-f 


A_y 


POLAR   MOMENT  OF  INERTIA    OF  PLANE   FIGURES.     Ill 


Hence,  summing,  and  passing  to  the  limit,  we  have 

ffx2dxdy  =  ffx2dxdy  +  2affXldxdy  +  a2  ffdxdy;   (i) 

but  if  we  were  seeking  the  abscissa  of  the  centre  of  gravity 
when  the  surface  is  referred  to  Yt  0  Ylt  and  if  this  abscissa  be 
denoted  by  x0,  we  should  have 


ffdxdy 

and,  since  x0  =  o,    .'.    ffx^dxdy  —  o;  hence,  substituting  this 
value  in  (i),  we  obtain 

ff'x2dxdy  =  ffx2dxdy  +  a2  ffdxdy.  (2) 

If,  now,  we  call  the  moment  of  inertia  about  OY,  7,  that 
about  O,  Ylt  7X,  and  let  the  area  —  A  —  ffdxdy,  we  shall  have 

/  =  7,  +  *^.  (3) 

Q.  E.  D. 

§93.  Polar  Moment  of  Inertia  of  Plane  Figures. —  The 

moment  of  inertia  of  a  plane 
figure  about  an  axis  perpen- 
dicular to  the  plane  is  equal 
to  the  sum  of  its  moments 
of  inertia  about  any  pair  of 
rectangular  axes  in  its  plane 
passing  through  the  foot  of 
the  perpendicular. 

PROOF.  —  Let  BCD  (Fig. 
63)  be  the  surface,  and  P  an 
elementary  area,  and  let 

OA  —  x,  AP  —  y,  OP  —  r;    then  the  moment  of   inertia  of 
the  surface  about  OZ  will  be 


ffr2dxdy 
Q.  E.  D. 


ff(x2 


=  ffx2dxdy  +  ffy2dxdy  = 


112  APPLIED   MECHANICS. 

Hence  follows,  also,  that  the  sum  of  the  moments  of  inertia 
of  a  plane  surface  relatively  to  a  pair  of  rectangular  axes  in  its 
own  plane  is  isotropic  ;  i.e.,  the  same  as  for  any  other  pair  of 
rectangular  axes  meeting  at  the  same  point,  and  lying  in  its 
plane. 

EXAMPLES. 

i.  To  find  the  moment  of  inertia  of  the  rectangle  (Fig.  59)  about 
an  axis  through  its  centre  perpendicular  to  the  plane  of  the  rectangle. 

Solution. 
Moment  of  inertia  about  YY  =  — , 

12 

Moment  of  inertia  about  an  axis  through  its 


hence 


centre  and  perpendicular  to  YY  —  — ; 

12 


bfc       hfr       bh ,,  -       LI 
Polar  moment  of  inertia  = \ =  — (h2  -f  b2). 

12  12  12 


2.  To  find  the  moment  of  inertia  of  a  circle  about  an  axis  through 
its  centre  and  perpendicular  to  its  plane  ( Fig.  60) . 

Solution. 

Moment  of  inertia  about  OY  =  — , 

4 

Moment  of  inertia  about  OX  =  —  ; 

4 
hence 

_  .,  Trr*         -n-r4         irr* 

Polar  moment  of  inertia  = \ =  — . 

442 

3.  To  find  the  moment  of  inertia  of  an  ellipse  about  an  axis  passing 
through  its  centre  and  perpendicular  to  its  plane. 


MOMENTS   OF  INERTIA    ABOUT  DIFFERENT  AXES. 


Solution. 
From  example  4,  §  91,  we  have 

,  TTdfc 

4 


.-.     Polar  moment  of  inertia  = (a2  -f  £2). 

4 


§  94.  Moments  of  Inertia  of  Plane  Figures  about  Different 
Axes  compared.  —  Given  the  surface  KLM  (Fig.  64),  suppose 
we  have  already  determined  the  quantities 

/  =  fftfdxdy,  J  =  fffdxdy,   K  =  ffxydxdy, 
it  is  required  to  determine,  in  terms  of  them,  the  quantities 
/i  =  ffxfdxjyi,  /,  =  ffyl*dxldyl,  K,  =  ffx^dx^y, ; 


the  angles  XOYznd  XlOYl  being  both 
right  angles,  and  YO  Yl  =  a. 

We  shall  have,  from  the  ordinary 
equations  for  the  transformation  .of  co- 
ordinates, to  be  found  in  any  analytic 
geometry,  the  equations 


xl  =  x  cos  a  +  y  sin  a, 
_y,  =  y  cos  a  —  x  sin  a, 
Tj2  =  .#2  cos2  a  -f-  _y2  sin2  a  +  2.#y  cos  a  sin  a, 


FIG.  64. 


,2  =  jc2  sin2  a 


cos2  a  —  2JTF  cos  a  sin  a, 


"ifi  =  xy  (cos2  a  —  sin2  a)  —  (x2  —  y2)  cos  a  sin  a. 


114  APPLIED   MECHANICS. 


Hence 


7X  =  ffx^dx.dy,  —  limit  of  ^xf&A 

=  cos2  a  limit  of  *Zx2&A  +  sin2  a  limit  of 

2  cos  a  sin  a  limit  of  ^xy^A 
=  (cQs2a.)ffx2dxdy  +  (sm2a)ffy2dxd'y  + 

2  (cos  a  sin  a)  ffxydxdy. 
/;  =  f'fyl2dxldyl  =  limit  of  Sj^A^ 

=  (sin2  a)    limit  of  2*2A,4  +  (cos2  a)   limit  of 

2  (cos  a  sin  a)  limit  of  ^xy^A 
=  ($itfa)ffx*dxdy  +  (cos2  a}ffy2dxdy  - 

2  (cos  a  sin  a)  ffxydxdy. 
,  =  ffx.y.dx.dy,  =  limit  of  ^x.y^A 
=  (cos2  a  —  sin2  a)  limit  of  ^xy^A  —  (cos  a  sin  a)   \  limit  of 

^x2^A  -  limit  of  2y2&A\ 
=  (cos2  a  —  sin2  a)  ffxydxdy  —  (cos  a  sin  a)  \ffx2dxdy  — 


Or,  introducing  the  letters  /,/,  and  ^T,  we  have 

/!  =  /cos2  a  +  y  sin2  a  -f-  2  A'  cos  a  sin  a,  (i) 

y,  =  /sin2  a  +  ycos2  a  —  2^  cos  a  sin  a,  (2) 

K,  =  (y—  /)  cos  a  sin  a  -f  ^(cos2a  —  sin2  a).  (3) 

The  equations   (i),   (2),  and  (3)   furnish   the   solution   of    the 
problem. 

§95.  Principal  Moments  of  Inertia  in  a  Plane.  —  In  every 
plane  figure,  a  given  point  being  assumed  as  origin,  there  is  at 
least  one  pair  of  rectangular  axes,  about  one  of  which  the  moment 
of  inertia  is  a  maximum,  and  a  minimum  about  the  other  ;  these 
moments  of  inertia  being  called  principal  moments  of  inertia, 
and  the  axes  about  which  they  are  taken  being  called  principal 
axes  of  inertia. 


AXES  OF  SYMMETRY  OF  PLANE   FIGURES.  11$ 

PROOF.  —  In  order  that  /,,  equation  (i),  §94,  may  be  a  maxi- 
mum or  a  minimum,  we  must  have,  as  will  be  seen  by  differen- 
tiating its  value,  and  putting  the  first  differential  co-efficient 
equal  to  zero, 

—  2/cos  a  sin  a  -h  2/cos  a  sin  a  +  2^T(cos2  a  —  sin2  a)  =  o 
.-.     K( cos2  a  —  sin2  a)  —  (/ — /)  cos  a  sin  a  =  o  (i) 

.-.         cosasina      =  _K_  -.     tan2a==_J^     (2) 

COS2  a  -  Sin2  a        7  -  /  /  -  / 

Hence,  for  the  value  of  a  given  by  (2),  we  have  7,  a  maximum 
or  a  minimum  ;  and  as  there  are  two  values  of  2a  corresponding 
to  the  same  value  of  tan  2a,  and  as  these  two  values  differ  by 
1 80°,  the  values  of  a  will  differ  by  90°,  one  corresponding  to  a 
maximum  and  the  other  to  a  minimum. 

Moreover,  when  the  value  of  a  is  so  chosen,  we  have 


as  is  proved  by  equation  (i).     Indeed,  we  might  say  that  the 
condition  for  determining  the  principal  axes  of  inertia  is 

K,  =  o. 

§96.  Axes  of  Symmetry  of  Plane  Figures.  —  An  axis 
which  divides  the  figure  symmetrically  is  always  a  principal 
axis. 

PROOF.  —  Let  us  assume  that  the  y  axis  divides  the  surface 
symmetrically  ;  then  we  shall  have,  with  reference  to  this  axis, 


And,  since  K  is  zero,  the  axis  of  y  is  one  principal  axis,  and  of 
course  the  axis  of  x  is  the  other.  The  same  method  of  reason- 
ing would  show  K  =  o  if  the  x  axis  were  the  axis  of  symmetry. 


1 1 6  APPLIED   MECHANICS. 

Hence,  whenever  a  plane  figure  has  an  axis  of  symmetry, 
this  axis  is  one  of  the  principal  axes,  and  the  other  is  at 
right  angles  to  it.  Thus,  for  a  rectangle,  when  the  axis  is  to 
pass  through  its  centre  of  gravity,  the  principal  axes  are  par- 
allel to  the  sides  respectively,  the  moment  of  inertia  being 
greatest  about  the  shortest  axis,  and  least  about  the  longest. 
Thus  in  an  ellipse  the  minor  axis  is  the  axis  of  maximum, 
and  the  major  that  of  minimum,  moment  of  inertia,  etc.  On 
the  other  hand,  in  a  circle,  or  in  a  square,  since  the  maximum 
and  minimum  are  equal,  it  follows  that  the  moments  of  inertia 
about  all  axes  passing  through  the  centre  are  the  same. 

§  97.  Conditions  for  Equal  Values  of  Moment  of  In- 
ertia. —  When  the  moments  of  inertia  of  a  plane  figure  about 
three  different  axes  passing  through  the  same  point  are  the 
same,  the  moments  of  inertia  about  all  axes  passing  through 
this  point  are  the  same. 

PROOF.  —  Let  /  be  the  moment  of  inertia  about  O  Y,  7, 
about  O  Ylt  I2  about  O  Y2,  and  let 

YOY,  =  a,         YOY2  =  ft 
and  let 

7,  =  72  =  7. 

Then,  from  equation  (i),  §94,  we  have 

/  =  /  cos2  a  +  J  sin2  a  -J-  2  K  cos  a  sin  a, 
7=  7  cos2/?  +/sin2/3  +  2  ^  cos/3  sinft 
Hence 

(7  — /)sin2a  =  2  A' cos  a  sin  a,  (i) 

(7-/)sm2/?=  2  AT  cos  £  sin  ft  (2) 

Hence 

(7-/)  tan  a  =  2K,  (3) 

(7-/)tan/2=  2*.  (4) 

And,  since  tan  a  is  not  equal  to  tan  ft  we  must  have 
7  —  J  =  o         and         K  =  o. 

Hence,  since  K  —  o  and  7  =J,  we  shall  have,  from  equa- 


MOMENTS   OF  INERTIA    ABOUT  PARALLEL   AXES.        1  1/ 

tion    (i),   §94,  for  the  moment   of   inertia  I'  about   an   axis, 
making  any  angle  0  with  O  Y, 

I'  =  /cos2  6  +  /sin2  6  +  o  =  /.  (5) 

Hence  all  the  moments  of  inertia  are  equal. 

§  98.  Components  of  Moments  of  Inertia  of  Solid 
Bodies.  —  Refer  the  body  to  three  rectangular  axes,  OX,  OY, 
and  OZ  ;  and  let  Ix,  Iy,  and  Iz  represent  its  moment  of  inertia 
about  each  axis  respectively.  Then,  if  r  denote  the  distance  of 
any  particle  from  OZt  we  shall  have 

/z  =  limit  of  *%wr2  ; 
but 

r*  =  x2  +  y2 

.'.    Iz  =  limit  of  %w(x2  +  y2)  =  limit  of  Sow2  +  limit  of  *2wy2.  (i) 
In  the  same  way  we  have 

Ix  =  limit  of  ^wy2  +  limit  of  2o/z2,  (2) 

Iy   —  limit  of  ^wx2  -h  limit  of  ^wz2.  (3) 

§99.  Moments  of  Inertia  of  Solids  around  Parallel 
Axes.  —  The  moment  of  inertia  of  a  solid  body  about  an  axis 
not  passing  through  its  centre  of  gravity  is  equal  to  its  moment 
of  inertia  about  a  parallel  axis  passing  through  the  centre  of 

gravity,  increased  by  the  product  of  the  entire  weight  of  the 
body  by  the  square  of  the  distance  between  the  two  axes. 

PROOF.  —  Refer  the  body  to  a  system  of  three  rectangular 
axes,  OX,  OY,  and  OZ,  of  which  OZ  is  the  one  about  which 
the  moment  of  inertia  is  taken.  Let  the  co-ordinates  of  the 
centre  of  gravity  of  the  body  with  reference  to  these  axes  be 
(*«  y*  *0).  Through  the  centre  of  gravity  of  the  body  draw  a 
system  of  rectangular  axes,  parallel  respectively  to  OX,  OY,  and 
OZ.  Then  we  shall  have  for  the  co-ordinates  of  any  point 


y  = 

z  = 


APPLIED   MECHANICS. 


Hence  v ' 

Iz  —  limit  of  ^w(x2  4-  y2)  —  limit  of  ^wx2  +  limit  of 
=  limit  of  ^w(x0  +  x,}2  -^  limit  of  *%w(y0  +  y,}2 
~  x02  limit  of  ^w  4-  j'02  limit  of  Hw  -f  2je0  limit  of 

4-  2>;0  limit  of  Sze/)^  4-  limit  of  ^wxf  4-  limit  of 
=  (^02  4-  _y02)  W  +  2jc0  limit  of  2^^!  4-  2j0  limit  of 

4-  limit  of  ^wrf 
=  r2W  +  //  4-  2.r0  limit  of  Sz^  -f-  2^0  limit  of 

But,  since  6^,  is  the  centre  of  gravity, 

/.     '^twxI  =  o         and         '^wyl  =  o. 
Hence 

/,  =  //  4.   Wr02, 

which  proves  the  proposition. 

§  100.   Examples  of  Moments  of  Inertia. 

i .  To  find  the  moment  of  inertia  of  a  sphere  whose  radius  is  r  and 
weight  per  unit  of  volume  w,  about  the  axis  OZ  drawn  through  its  centre. 

Solution. 

Divide  the  sphere  into  thin  slices  (Fig.  65)  by  planes  drawn  perpen- 
dicular to  OZ.  Let  the  distance 
of  the  slice  shown  in  the  figure, 
above  O  be  z,  and  its  thickness  dz : 
then  will  its  radius  be  vV2  —  z2 ; 
and  we  can  readily  see,  from  ex- 
ample 2,  §  93,  that  its  moment  of 
inertia  about  OZ  will  be 


—  z2) 


dz. 


FIG.  65. 


Hence  the  moment  of  inertia 
of  the  entire  sphere  about  OZ  will 
be 


-   \     (r>  -  z2)2dz, 

2  U—  r 


EXAMPLES   OF  MOMENTS   OF  INERTIA. 


IIQ 


which  easily  reduces  to 


2.  To  find  the  moment  of  inertia  of  an  ellipsoid  (semi-axes  a,  t>,  c) 
about  OZ  (Fig.  66). 

SOLUTION. — The  equa- 
tion of  the  ellipsoid  is 


a*       P     •£* 

Divide  it  into  thin  slices 
perpendicular  to  OZ,  and 
let  the  slice  shown  in  the 
figure  be  at  a  distance  z 
from  O.  Then  will  this 
slice  be  elliptical,  and  its 
semi-axes  will  be 

a   i 


66. 


and 


and  from  example  3,  §  93,  we  readily  obtain,  for  its  moment  of  inertia 
about  OZ, 


Hence,  for  the  moment  of  inertia  of  the  ellipsoid  about   OZ,  we 


have 


(a*  4-  b2}   Cc 
4^4  J-  c 


T5 


a*  4- 


3.  Find  the  moment  of  inertia  of  a  right  circular  cylinder,  length  a, 
radius  r,  about  its  axis. 

Ans. 


I2O  APPLIED   MECHANICS. 

4.    Find  the  moment  of  inertia  of  the  same  about  an  axis  perpen- 
dicular to,  and  bisecting  its  axis. 


4      V  3 

5.    Find  the  moment  of  inertia  of  an  elliptic  right  cylinder,  length 
2c,  transverse  semi-axes  a  and  b,  about  its  longitudinal  axis. 


.  ,  -        7,x 

Ans.  —\a    +  *  )• 

2 

6.   Find  the  moment  of  inertia  of  the  same  about  its  transverse 


\4         3 

7.  Find  the  moment  of  inertia  of  a  rectangular  prism,  sides  za,  2b, 
2C,  about  central  axis  2C.  Ans.     \wabe(&  -f-  b2). 

§  101.    Centre  of   Percussion.  —  Suppose  we  have  a  body 
revolving,  with  an  angular  velocity  a,  about  an  axis  perpendicu- 

lar to  the  plane  of  the  paper,  and 
passing  through  O.  Join  O  with 
the  centre  of  gravity,  G,  and  take  OG 
as  -axis  of  x;  the  axis  of  y  passing 
through  O,  and  lying  in  the  plane  of 
the  paper.  If,  with  a  radius  OA  =  r, 
we  describe  an  arc  CA  (Fig.  67),  all 
particles  situated  in  this  arc  have  a 

linear  velocity  a.r.     The  force  which  would  impart  this  velocity 
to  any  one  of  them,  as  that  at  A,  in  a  unit  of  time,  is 

Ear, 

g 

and  this  may  be  resolved  into  two, 

w  t     w 

—ax      and      —ay, 

g  g 

respectively  perpendicular  and  parallel  to  OG.    The  moment  of 
this  force  about  the  axis  is 


8 
hence  the  total  moment  of  the  forces  which  would  impart  to 


CENTRE   OF  PERCUSSION.  121 


the  body  in  a  unit  of  time  the  angular  velocity  a,  is,  as  has  been 
shown  already, 


g        g 
The  resultant  of  the  forces  acting  on  the  body  is 


-Sow, 

g 

since,   the   centre   of   gravity   being   on   OB,    it   follows   that 
^wy  •=•  o ;  and  hence 

g 

Hence  the  perpendicular  distance  from  O  to  the  line  of  direc- 
tion of  the  resultant  force  is  measured  along  OG,  and  is 


and  the  point  of  application  of  the  resultant  force  may  be  con- 
ceived to  be  at  a  point  on  OG  at  a  distance  /  from  O  ;  and  this 
point  of  application  of  the  resultant  of  the  forces  which  pro- 
duce the  rotation  is  called  the  Centre  of  Percussion. 

If  p  =  radius  of  gyration  about  the  axis  through  O,  and  if 
x0  =  distance  from  O  to  the  centre  of  gravity,  we  have 


Hence 


X 


or,  in  words,  — 

The  radius  of  gyration  is  a  mean  proportional  between  the 
distance  1,  and  the  distance  x0,  between  the  axis  of  oscillation  and 
the  centre  of  gravity. 

The  centre  of  percussion  with  respect  to  a  given  axis  of 
oscillation  O  has  been  defined  as  the  point  of  application  of  the 


122  APPLIED   MECHANICS. 

resultant  of  the  forces  which  cause  the  body  to  rotate  around  the 
point  O. 

Another  definition  often  given  is,  that  it  is  the  po int  at  which, 
if  a  force  be  applied,  there  will  be  no  shock  on  the  axis  of  oscilla- 
tion ;  and  these  two  definitions  are  equivalent  to  each  other. 

Let  the  particles  of  the  body  under  consideration  be  con- 
ceived, for  the  sake  of  simplicity,  to  be  distributed  along  a  single 
line  AB,  and  suppose  a  force  F  applied  at  D 
(Fig.  68).  Conceive  two  equal  and  opposite 
forces,  each  equal  to  F,  applied  at  C,  the  cen- 
tre of  gravity  of  the  body. 

3 Then  these  three  forces  are  equivalent  to 

a  single  force  .F  applied  at  the  centre  of  grav- 
ity C,  which  produces  translation  of  the  whole 
body ;  and,  secondly,  a  couple  whose  moment 
B  is  F(CD),  whose  effect  is  to  produce  rotation 

FIG.  68.  around  an  axis  passing  through  the  centre  of 

gravity  C.     Under  this  condition  of  things  it  is  evident  that  the 

centre  of  gravity  C  will  have  imparted  to  it  in  a  unit  of  time  a 

77 
forward  velocity  equal  to  — ,  where  M  is  the  entire  mass  of  the 

body  ;  the  point  D  will  have  imparted  to  it  a  greater  forward 
velocity ;  while  those  points  on  the  upper  side  of  C  will  have 
imparted  to  them  a  less  and  less  velocity  as  they  recede  from 
C,  until,  if  the  rod  is  sufficiently  long,  the  particle  at  A  will 
acquire  a  backward  velocity. 

Hence  there  must  be  some  point  which  for  the  instant  in 
question  is  at  rest;  i.e.,  where  the  velocity  due  to  rotation  is  just 
equal  and  opposite  to  that  due  to  the  translation,  or  about 
which,  for  the  instant,  the  body  is  rotating:  and  if  this  point 
were  fixed  -by  a  pivot,  there  would  be  no  stress  on  the  pivot 
caused  by  the  force  applied  at  D. 

An  axis  through  this  point  is  called  the  Instantaneous 
Axis. 


IMPACT  OR   COLLISION.  123 

§  102.  Intel-changeability  of  the  Centre  of  Percussion 
and  Centre  of  Oscillation.  —  If  we  take  our  centre  of  percus- 
sion D  as  axis  of  oscillation,  then  will  O  be  the  new  centre  of 
percussion. 

PROOF.  —  We  have  seen  (§  101)  that 

'•5 

where  /  =  OD,  XQ  =  OC,  and  P  —  radius  of  gyration  about  an 
axis  through  O  perpendicular  to  the  plane  of  the  paper. 

Moreover,  if  />0  represent  the  radius  of  gyration  about  an 
axis  through  C  perpendicular  to  the  plane  of  the  paper,  we  shall 
have 

P2  =  Pa2  +  *02 


Now  if  D  is  taken  as  axis  of  oscillation,  we  shall  have  for  the 
distance  /z  to  the  corresponding  centre  of  percussion, 


CD   i  -  x: 

where  p,  =  radius   of   gyration  about  the   axis   of   oscillation 
through  D. 

.         _     Pl»  _  Po»  +  CD*  __   po2     ,    CD-x 
~~CD~      ~CD~      ~~* 


Hence  the  new  centre  of  percussion  is  at  O.     Q.  E.  D. 

§  103.  Impact  or  Collision.  —  Impact  or  collision  is  a 
pressure  of  inappreciably  short  duration  between  two  bodies. 

The  direction  of  the  force  of  impact  is  along  the  straight  line 
drawn  normal  to  the  surfaces  of  the  colliding  bodies  at  their 
point  of  contact,  and  we  may  call  this  line  the  line  of  impact. 


124  APPLIED   MECHANICS. 

The  action  that  occurs  in  the  case  of  collision  may  be  de- 
scribed as  follows  :  at  first  the  bodies  undergo  compression  ; 
the  mutual  pressure  between  them  constantly  increasing,  until, 
when  it  has  reached  its  maximum,  the  elasticity  of  the  mate- 
rials begins  to  overpower  the  compressive  force,  and  restore 
the  bodies  wholly  or  partially  to  their  original  shape  and  dimen- 
sions. 

Central  impact  occurs  when  the  line  joining  the  centres  of 
gravity  of  the  bodies  coincides  with  the  line  of  impact. 

Eccentric  impact  occurs  when  these  lines  do  not  coincide. 

Direct  impact  occurs  when  the  line  along  which  the  relative 
motion  of  the  bodies  takes  place,  coincides  with  the  line  of 
impact. 

Oblique  impact  occurs  when  these  lines  do  not  coincide. 

CENTRAL    IMPACT. 

§  104.  Equality  of  Action  and  Re-action.  —  One  funda- 
mental principle  that  holds  in  all  cases  of  central  impact  is  the 
equality  of  action  and  re-action ;  in  other  words,  we  must  have, 
that,  at  every  instant  of  the  time  during-which  the  impact  is 
taking  place,  the  pressure  that  one  body  exerts  upon  the  other 
is  equal  and  opposite  to  that  exerted  by  the  second  upon  the 
first. 

The  direct  consequence  of  this  principle  is,  that  the  algebraic 
sum  of  the  momenta  of  the  two  bodies  before  impact  remains 
unaltered  by  the  impact,  and  hence  that  this  sum  is  just  the 
same  at  every  instant  of,  and  after,  the  impact. 

If  we  let 

m»  m2,  be  the  respective  masses, 
£„    c2,  their  respective  velocities  before  impact, 

z>M   VM  their  respective  velocities  after  impact, 

z/,   v",  their  respective  velocities  at  any  given  instant  during 
the  time  while  impact  is  taking  place, 


CO-EFFICIENT  OF  RESTITUTION.  125 

then  we  must  have  the  following  two  equations  true  ;  viz.,  — 
m^v,  +  m2v2  =  m^cl  -f  m2c2,  (i) 

mtvr  -|-  m2v"  —  mlcl  -\-  m2c2.  (2) 

§  105.  Velocity  at  Time  of  Greatest  Compression.  —  At 
the  instant  when  the  compression  is  greatest  —  i.e.,  at  the 
instant  when  the  elasticity  of  the  bodies  begins  to  overcome 
the  deformation  due  to  the  impact,  and  to  tend  to  restore  them 
to  their  original  forms  —  the  values  of  i/  and  v"  must  be  equal 
to  each  other;  in  other  words,  the  colliding  bodies  must  be 
moving  with  a  common  velocity 

v  =  if  =  if'.  (i) 

To  determine  this  velocity,  we  have,  from  equation  (2),  §  104, 
combined  with  (i), 

^  __  IH^CI  -f-  ^2^2  /2\ 


§  1  06.  Co-efficient  of  Restitution.  —  In  order  to  determine 
the  values  vlt  v2,  of  the  velocities  after  impact,  we  need  two 
equations,  and  hence  two  conditions.  One  of  them  is  fur- 
nished by  equation  (i),  §  104.  The  second  depends  upon  the 
nature  of  the  material  of  the  colliding  bodies,  and  we  may  dis- 
tinguish three  cases  :  — 

i°.  Inelastic  Impact.  —  In  this  case  the  velocity  lost  up  to 
the  time  of  greatest  compression  is  not  regained  at  all,  and 
the  velocity  after  impact  is  the  common  velocity^  at  the  instant 
of  greatest  compression.  In  this  case  the  whole  of  the  work 
used  up  in  compressing  the  bodies  is  lost,  as  none  of  it  is 
restored  by  the  elasticity  of  the  material. 

2°.  Elastic  Impact.  —  In  this  case  the  velocity  regained 
after  the  greatest  compression,  is  equal  and  opposite  to  that 
lost  up  to  the  time  of  greatest  compression  ;  therefore 

v  —  vI  =  d  —  v.       (i)  v2  —  v  =  v  —  c2.     (2) 


126  APPLIED   MECHANICS. 

We  may  also  define  this  case  as  that  in  which  the  work  lost 
in  compressing  the  bodies  is  entirely  restored  by  the  elasticity 
of  the  material,  so  that 

=  ^!  +  ^.  (3) 


Either  condition  will  lead  to  the  same  result. 

3°.  Imperfectly  Elastic  Impact.  —  In  this  case  a  part  only 
of  the  velocity  lost  up  to  the  time  of  greatest  compression  is 
regained  after  that  time. 

If,  when  the  two  bodies  are  of  the  same  material,  we  call  e 
the  co-efficient  of  restitution,  then  we  shall  so  define  it  that 

v  —  Vi 


—   V          V   —   C 


or,  in  words,  the  co-efficient  of  restitution  is  the  ratio  of  the 
velocity  regained  after  compression  to  that  lost  previous  to 
that  time. 

In  this  case  only  a  part  of  the  work  done  in  producing  the 
compression  is  regained,  hence  there  is  loss  of  energy.  Its 
amount  will  be  determined  later. 

Strictly  speaking,  all  bodies  belong  to  the  third  class  ;  the 
value  of  e  being  always  a  proper  fraction,  and  never  reaching 
unity,  the  value  corresponding  to  perfect  elasticity  ;  nor  zero, 
the  value  corresponding  to  entire  lack  of  elasticity. 

§  107.  Inelastic  Impact.  —  In  this  case  the  velocity  after 
impact  is  the  common  velocity  at  the  time  of  greatest  com- 
pression ;  hence 

V  =   V,   =   V2  (l) 


ml  -j-  m2 
And  for  the  loss  of  energy  due  to  impact  we  have 


ELASTIC  IMPACT.  1 27 


which,  on  substituting  the  value  of  v,  reduces  to 

;i?-   ;       .(„!+' *.) (Ci  ~  C*Y-          (3) 

§  1 08.  Elastic  Impact.  —  In  this  case  we  have,  of  course, 

the  condition,  equation  (i),  §  104, 


mlvl  +  m2v2  =  mlcl  -f- 
and,  for  second  equation,  we  may  use  equation  (3),  §  106  ;  viz., 


2222 

Combining  these  two  equations,  we  shall  obtain 


2m,(cl  —  A)  /  x 

"-!  -  a*.  (2) 


We  can  obtain  the  same  result  without  having  to  solve  an 
equation  of  the  second  degree,  by  using  instead  the  equations 
(i)  and«(2)  of  §  106,  together  with  (i)  of  §  104;  i.e.,  — 

nijVi     +     ^2^2     =     Wl^I     + 

v  —  2>t  =  Ci  —  v, 
or 

v2  -  v  =  v  -  czy 
and  (§  105) 


As  the  result  of  combining  these  equations,  and  eliminating 
v,  we  should  obtain  equations  (i)  and  (2),  as  above,  for  the  values 
of  vl  and  v2.  In  this  case  the  energy  lost  by  the  collision  is 
zero. 


128  APPLIED   MECHANICS. 

§  109.   Special  Cases  of  Inelastic  Impact.  —  (a)   Let  the 
mass  m2  be  at  rest.     Then  c2  =  o, 


/  .     Loss  of  energy  =         iz     -±-.  (2) 

ml  +  m2  2 


(b)  Let  m2  be  at  rest,  and  let  m2  —  oo  ;  i.e.,  let  the  mass  m, 
strike  against  another  which  is  at  rest,  and  whose  mass  is  in- 
finite. We  have 

m2  =  co ,         c2  =  o, 


f  ^ 

=   O,  (3) 


m2 

t  r  i  !  /     x 

Loss  of  energy  =  --  =  -  ,  (4) 


or  the  moving  body  is  reduced  to  rest  by  the  collision,  and  all 
its  energy  is  expended  in  compression. 

(c)  Let  mlcl  =  —m2c2  ;  i.e.,  let  the  two  bodies  move  towards 
each  other  with  equal  momenta  : 


m         m 


J         ,  1  C  ?#I^I2        ,        ^2^22  f£.\ 

and  the  loss  of  energy  =  —  —  —  f-  •        ,  (v) 

the  entire  energy  being  lost. 

§110.   Special    Cases    of    Elastic    Impact.  —  (a)  Let  the 
mass  m2  be  at  rest.     Then  c2  =  o, 

2.m2c  ',  /  \ 

v,.  =  c,  --  ^—  (i) 

m^  -t-  mz 

2mlcI  /  \ 

s.    v2  =  —  -L-L~.  (2) 

m,  -f-  mz 


EXAMPLES   OF  ELASTIC  AND   INELASTIC  IMPACT.       I2Q 

(b)  Let  m2  be  at  rest,   and  let  m2  —  oo .     Then  we  have 
r2  =  o, 

c,  -  2f,  =  -<:„  (3) 


V2   =   O.  (4) 

Hence  the  moving  body  retraces  its  path  in  the  opposite  direc- 
tion with  the  same  velocity. 

(c)  Let  m1cl  =  —  m2c2.     Then    our  equations    of   condition 
become 

mlvl  -h  m2vz  =  o, 

m^2    .    m2v2z        m^c2    .    m2cz2 

--  1  --  =  --  1  --  : 

2222 

and  from  these  we  readily  obtain 


i.e.,  both  bodies  return  on  their  path  with  the  same  velocity 
with  which  they  approached  each  other. 

§m.  Examples  of  Elastic  and  of  Inelastic  Impact. 

1.  With  what  velocity  must  a  body  weighing  8  pounds  strike  one 
weighing  25  pounds  in  order  to  communicate  to  it  a  velocity  of  2  feet 
per  second,  (a)  when  the  bodies  are  perfectly  elastic,  (b)  when  wholly 
inelastic. 

2.  Suppose  sixteen  impacts  per  minute  take  place  between  two  bodies 
whose  weights  are  respectively  1000  and  1200  pounds,  their  initial  velo- 
cities being  5  and  2  feet  per  second  respectively  :  find  the  loss  of  energy, 
the  bodies  being  inelastic. 

§112.  Imperfect   Elasticity.  —  In  this  case  we  have  the 
relations  (see  §  106) 


V  — 


Vj.   _   V2    —   V  _ 


l    —   V          V   — 


130  APPLIED   MECHANICS. 

where 

i 

?H  ±C  i    -f- 


mt  4-  «a 
and  we  have  also 

#2l#I     4~     ^2^2     = 

Determining  from  them  the  values  of  v^  and  v2>  we  obtain 
v,  =  v(i  4-  e)  —  eclt  (i) 

v2  =  »(i  4-  0  -  <*„  (2) 

or,  by  substituting  for  #  its  value, 

p- = '";«:  +  "':!* {i+e)- ec"  -    .  <*> 
;  ^ =  mm,  + '"? (i  +  e)-  ec>-  •  :  w 

These  may  otherwise  be  put  in  the  form 

^=^-(14-  0  jj^jj-  (^  -  cj,        (5) 

»a  =  c2  4-  (i  4-  ^)  — ^ (/,  -  O-        (6) 

m,  4-  /«2 

Moreover,  we  have  for  the  loss  of  energy  due  to  impact 

2  2 

or 

but,  from  (5)  and  (6)  respectively, 

^!      —     Vl     =     J 


m,  4-  m2 

4-  e}ml^cl  - 
ml  4-  m2 


IMPERFECT  ELASTICITY.  1  3  I 


But,  from  (i)  and  (2), 


E  = 


or 

£=  (i-  <•)  2(Jl™2m}  ('.  -  O2-  (7) 

When  e  •=.  i,  or  the  elasticity  is  perfect,  this  loss  of  energy 
becomes  zero. 

When  e  ==  o,  or  the  bodies  are  totally  inelastic,  then  the  loss 
of  energy  becomes 


,  , 

2(01,  +  m2) 


(8) 


as  has  been  already  shown  in  §  107. 

An  interesting  fact  in  this  connection  is,  that  since  (8)  is 
the  work  expended  in  producing  compression,  and  (7)  is  the 
work  lost  in  all,  therefore  the  work  restored  by  the  elasticity  of 
the  body  is 

2  ( ml  +  *#2 

so  that  ^2,  or  the  square  of  the  co-efficient  of  restitution,  is  the 
ratio  of  the  work  restored  by  the  elasticity  of  the  bodies,  to 
the  work  expended  in  compressing  the  bodies  up  to  the  time 
of  greatest  compression. 


132  APPLIED   MECHANICS. 

§113.    Special  Cases.  —  (a)    Let  m2  be  at  rest,  therefore 
c2  =  o.     Then  we  shall  have 


and  for  loss  of  energy 

-  ^^  -  c*.  (3) 


(b)  When  m2  =  oo  ,  and  ^2  =  o,  we  have 

»i  =  -«•.,  (4) 


(s) 


When  m,cl  =  —m2c2)  then 

v,  =  —ect 


2m  2 


§114.    Values   of    e   as    Determined   by   Experiment. 
Since  we  have 


<:T  —  V 


IMPERFECT  ELASTICITY.  133 

we  shall  have,  when 

m2  =  oo      and      cz  =  o, 

v  =  m^  +  ;^2  =  o. 
01,  H-  z#2 

Hence 


Now,  if  we  let  a  round  ball  fall  vertically  upon  a  horizontal 
slab  from  the  height  H,  we  shall  have  for  the  velocity  of  ap- 
proach 


and  if  we  measure  the  height  h  to  which  it  rises  on  its  rebound, 
we  shall  have 


Hence 

In  this  way  the  value  of  e  can  be  determined  experimentally 
for  different  substances.  ' 

Newton  found  for  values  of  e:  for  glass,  ^| ;  for  steel,  |; 
and  Coriolis  gives  for  ivory  from  0.5  to  0.6. 

On  the  other  hand,  if  we  desired  to  adopt  as  our  constant 
the  ratio  of  the  work  restored,  to  the  work  spent  in  compres- 
sion, we  should  have  for  our  constant  ^,  and  hence  the  squares 
of  the  preceding  numbers. 

EXAMPLES. 

i.  If  two  trains  of  cars,  weighing  120000  and  160000  Ibs.,  come 
into  collision  when  they  are  moving  in  opposite  directions  with  veloci- 
ties 20  and  15  feet  per  second  respectively,  what  is  the  loss  of  mechan- 
ical effect  expended  in  destroying  the  locomotives  and  cars  ? 


134  APPLIED  MECHANICS. 

2.  Two  perfectly  inelastic   balls   approach   each   other  with   equal 
velocities,  and  are  reduced  to  rest  by  the  collision ;  what  must  be  the 
ratio  of  their  weights  ? 

3.  Two  steel  balls,  weighing  10  Ibs.  each,  are  moving  with  velocities 
5  and  10  feet  per  second  respectively,  and  in  the  same  direction  :  find 
their  velocities  after  impact,  the  fastest  ball  being  in  the  rear,  and  over- 
taking the  other ;  also  the  loss  of  mechanical  effect  due  to  the  impact,, 
assuming  e  —  0.55. 

§115.  Oblique  Impact. 

Let  miy  m2,  be  the  masses  of  the  colliding  bodies ; 
cu  c2,  their  respective  velocities  before  impact ; 
a,,  a2,  the  angles  made  by  clt  c2t  with  the  line  of  centres ; 
vlt  v2,  the  components  of  the  velocities  after  impact ; 
^cosctj,  <:2cosa2,  the  components  of  clt  c2t  along  the  line  of 

centres  ; 
cl  sin  alt  c2  sin  a2,  the  components  of  cu  c2t  at  right  angles  to 

the  line  of  centres  ; 
v  the  common  component  of  the  velocity  at  the  instant  of 

greatest  compression  along  line  of  centres ; 
z/,  v",  actual  velocities  after  impact ; 
a',  a",  angles  they  make  with  line  of  centres  ; 
vcr,  v" y  actual  velocities  when  compression  is  greatest ; 
a/,  a/',  angles  they  make  with  line  of  centres. 

Then  we  shall  have,  by  proceeding  in  the  same  way  as  was  done 
in  §  112, 

vl  =  r.cosa,  —  (i  H-  e) — — (/J  cos  a,  —  r2cosa2),    (i) 

ml  +  m2 


^2COSa2),     (2) 

l     +     0*2 


OBLIQUE   IMPACT.  135 


if  =  V^2  H-  ^sin'a,,  (3) 


i/'  =  V^22  H-  cf  sin2  a,,  (4) 

cosa'  =  ^,  (5) 

(6) 

v 

/«,^ICOSaI  -f-  ^2^2  COS  03  ,    x 

# • ,  17; 


(8) 
(9) 

COS  a/  =  ~,  (lo) 


COSa/  =  -£. 


And  for  the  energy  lost  in  impact,  we  have 

E  =  (i  -  *2) ^^ (^cosox  -  ;acosa,> 

When  the  bodies  are  perfectly  elastic, 

e  =  i, 

and  equations  (i),  (2),  and  (12)  become  respectively 

2m?      ,  x 

z/x  =  ^TJ  cos  a, — (V,  cos  ax  —  c2  cos  03), 

V2  =  ^2  COS  a2  H ^5 (<;j  cos  aj  _  ^  cos  Ojj), 

jE1  =  o. 

The  rest  remain  the  same  in  form. 

When  the  bodies  are  totally  inelastic, 

e  =  o, 


136  APPLIED   MECHANICS. 

and  equations  (i),  (2),  and  (12)  become  respectively 

V^  =  Cj  COS  ax   --  —  -  ( 


V2  =  <r2  cos  03  H  --  —  -  (^  cos  c^  —  <r2  cos 
/«,  4-  w2 


2(ms  4-  «3) 
The  rest  remain  the  same  in  form.        ~  / 

§116.   Impact   of   Revolving   Bodies.  —  Let  the  bodies  A 
and  B  revolve  about  parallel  axes,  and  impinge  upon  each  other. 
Draw  a  common   normal  at  the  point  of  contact.      This 
common  normal  will  be  the  line  of  impact. 
Let  ex  =  angular  velocity  of  A  before  impact, 
€2  =  angular  velocity  of  B  before  impact, 
Wl  =:  angular  velocity  of  A  after  impact, 
u>2  =  angular  velocity  of  B  after  impact, 
al  =  perpendicular  from  axis  of  A  on  line  of  impact, 
a2  =  perpendicular  from  axis  of  B  on  line  of  impact, 
/r  =  moment  of  inertia  of  A  about  its  axis, 
7a  =  moment  of  inertia  of  B  about  its  axis. 

Then  we  shall  have 

tfl€r  =  d  =  linear  velocity  of  A  at  point  of  contact  before  impact  ; 

#2e2  =  c2  =  linear  velocity  of  B  at  point  of  contact  before  impact  ; 

al(DI  =  vs  =  linear  velocity  of  A  at  point  of  contact  after  impact  ; 

a2u2  =  v2  =  linear  velocity  of  B  at  point  of  contact  after  impact  ; 

72          //\/r2 

ILJ_  =  f—L  \_L.  =  actual  energy  of  A  before  impact  ; 
2g         \al2/2g 

^  =  /  A\£2!  =  actual  energy  of  B  before  impact  ; 
2g         \a22/2g 

^L.  —  I  £L\^L  =  actual  energy  of  A  after  impact  ; 
2g         \asj2g 

l^  =  (A\?l  -  actual  energy  of  B  after  impact  ; 

2g  W/  2g 


IMPACT  OF  REVOLVING   BODIES.  137 

Hence  it  follows  that  we  have  the  case  explained  in  §  112  for 
imperfectly  elastic  impact,  provided  only  we  write 

—  -  instead  of  m,g     and     —  instead  of  m2g. 
a*  a22 

Hence  we  shall  have 

a>x  =  e,  —  «,(«,€!  -  a2€2)  2        t  (i  4-  <?),       (i) 

/A  4-A^ 

ft),    =    £2     -f     02Oiei     -     «2€2)-  -  -J—  -  (i     4-    <?),  (2) 


The  case  of  perfect  elasticity  is  obtained  by  making  e  —  i. 
The  case  of  total  lack  of  elasticity  is  obtained  by  making 
e  —  o. 

In  the  latter  case  the  loss  of  energy  is       . 


XV 


as  can  be  seen  by  substituting  the  proper  values  in  equation  (8), 
§  U2. 


138  APPLIED   MECHANICS. 


CHAPTER   III. 
ROOF-TRUSSES. 

§  117.  Definitions  and  Remarks The  term  "truss"  may 

be  applied  to  any  framed  structure  intended  to  support  a  load. 

In  the  case  of  any  truss,  the  external  loads  may  be  applied 
only  at  the  joints,  or  some  of  the  truss  members  may  support 
loads  at  points  other  than  the  joints. 

In  the  latter  case  those  members  are  subjected,  not  merely 
to  direct  tension  or  compression,  but  also  to  a  bending-action, 
the  determination  of  which  we  shall  defer  until  we  have  studied 
the  mode  of  ascertaining  the  stresses  in  a  loaded  beam  ;  and 
we  shall  at  present  confine  ourselves  to  the  consideration  of 
the  direct  stresses  of  tension  and  compression. 

For  this  purpose  any  loads  applied  between  two  adjacent 
joints  must  be  resolved  into  two  parallel  components  acting  at 
those  joints,  and  the  truss  is  then  to  be  considered  as  loaded  at 
the  joints.  By  this  means  we  shall  obtain  the  entire  stresses  in 
the  members  whenever  the  loads  are  concentrated  at  the  joints ; 
and,  when  certain  members  are  loaded  at  other  points,  our  re- 
sults will  be  the  direct  tensions  and  compressions  of  these  mem- 
bers, leaving  the  stresses  due  to  bending  yet  to  be  determined. 

A  tie  is  a  member  suited  to  bear  only  tension. 

A  strut  is  a  member  suited  to  bear  compression. 

§  1 1 8.  Frames  of  Two  Bars.  —  Frames  of  two  bars  may 
consist,  (i)  of  two  ties  (Fig.  69),  (2)  of  two  struts  (Fig.  70), 
(3)  of  a  strut  and  a  tie  (Fig.  71). 


FRAMES   OF   TWO  BARS. 


139 


CASE   I.    Two    Ties   (Fig.  69).  —  Let   the   load   be   repre- 
sented graphically  by  CF  =  W.    a 
Then   if    we   resolve  it   into 
two  components,  CD  and  CE, 
acting  along  CB  and  CA  re- 
spectively, CD  will  represent 
graphically  the  pull  or  tension 
in  the  tie  CB,  and  CE  that  in 
the  tie  CA. 

The  force  acting  on  CB  at 
B  is  equal  and  opposite  to 
CD,  while  that  acting  on  CA  at  A  is  equal  and  opposite  to  CE. 

To  compute  these  stresses  analytically,  we  have 


FIG.  69. 


rp 


CD  =  CF 


sin  CFE 
sin  CEF 

sin  CFD 
sin  CDF 


sn 


W 


sin(/  4-  ' 

sin  /, 
sin(z  4- 


CASE  II.  Two  Struts  (Fig.  70).  —  Let  the  load  be  repre- 
sented graphically  by  CF—  W. 
Then  will  the  components  CD 
and  CE  represent  the  thrusts 
in  the  struts  CB  and  CA  re- 
spectively, and  the  re-actions 
of  the  supports  at  B  and  A 
will  be  equal  and  opposite  to 
them.  For  analytical  solution, 
we  derive  from  the  figure 


FIG.  70. 


CE  =  W 


sin  i, 


sin  (* 


CD  =  JF- 


smz 


sin(* 


CASE  III.  A  Strut  and  a  Tie  (Fig.  71).  —  Let  the  load  be 
represented  graphically  by  CF  =  W.  Resolve  it,  as  before, 
into  components  along  the  members  of  the  truss,  Then  will 


140  APPLIED   MECHANICS. 

CE  represent  the  tension  in  the  tie  AC,  and  CD  will  represent 
the  thrust  in  the  strut  BC ;  and  we  may 
deduce  the  analytical  formulae  as  before. 

§  1 19.  Stability  for  Lateral  Deviations. 
—  In  Case  I,  if  the  joint  C  be  moved  a  little 
out  of  the  plane  of  the  paper,  the  load  at 
C  has  such  a  direction  that  it  will  cause  the 
truss  to  rotate  around  AB  so  as  to  return  to 
its  former  position  ;  hence  such  a  frame  is 
stable  as  regards  lateral  deviations. 

In  Case  II    the  effect  of   the  load,  if  C 
were  moved  a  little  out  of  the  plane  of  the 
paper,  would  be  to  cause  rotation  in  such  a  way  as  to  overturn 
the  truss ;    hence  such  a  frame  is  unstable  as  regards  lateral 
deviations. 

In  Case  III  the  stability  for  lateral  deviations  will  depend 
upon  whether  the  load  CF  =  W  is  parallel  to  AB,  is  directed 
away  from  it  or  towards  it.  If  the  first  is  the  case  (i.e.,  if  A  is 
the  point  of  suspension  of  the  tie),  the  frame  is  neutral,  as  the 
load  has  no  effect,  either  to  restore  the  truss  to  its  former  posi- 
tion, or  to  overturn  it ;  if  the  second  is  the  case  (i.e.,  if  At  is 
the  point  of  suspension  of  the  tie),  the  truss  is  stable ;  and,  if 
the  third  is  the  case  (i.e.,  if  A2  is  the  point  of  suspension  of  the 
tie),  it  is  unstable  as  regards  lateral  deviations. 

§  1 20.  General  Methods  for  Determining  the  Stresses  in 
Trusses.  —  In  the  determination  of  the  stresses  as  above,  it 
would  have  been  sufficient  to  construct  only  the  triangle  CFD 
by  laying  off  CF—  Wio  scale,  and  then  drawing  CD  parallel 
to  CB,  and  FD  parallel  to  CA,  and  the  triangle  CFD  would  have 
given  us  the  complete  solution  of  the  problem.     Moreover,  the 
determination  of  the  supporting  forces  of  any  truss,  and  of  the 
stresses  in  the  several  members,  is  a  question  of  equilibrium. 
Adopting  the  following  as  definitions,  viz.,  — 
External  forces  are  the  loads  and  supporting  forces, 


TRIANGULAR  FRAME.  141 

Internal  forces  are  the  stresses  in  the  members  : 
we  must  have 

i°.  The  external  forces  must  form  a  balanced  system;  i.e., 
the  supporting  forces  must  balance  the  loads. 

2°.  The  forces  (external  and  internal)  acting  at  each  joint 
of  the  truss  must  form  a  balanced  system  ;  i.e.,  the  external 
forces  (if  any)  at  the  joint  must  be  balanced  by  the  stresses  in 
the  members  which  meet  at  that  joint. 

3°.  If  any  section  be  made,  dividing  the  truss  into  two  parts, 
the  external  forces  which  act  upon  that  part  which  lies  on  one 
side  of  the  section,  must  be  balanced  by  the  forces  (internal) 
exerted  by  that  part  of  the  truss  which  lies  on  the  other  side 
of  the  section,  upon  the  first  part. 

The  above  three  principles,  the  triangle,  and  polygon  of 
forces,  and  the  conditions  of  equilibrium  for  forces  in  a  plane, 
enable  us  to  determine  the  stresses  in  the  different  members 
of  roof  and  bridge  trusses. 

§121.  Triangular  Frame.  —  Given  the  triangular  frame 
ABC  (Fig.  72),  and  given  the  load  W  at  C  in  magnitude  and 
direction,  given  also  the 
direction  of  the  support- 
ing  force  at  B,  to  find  the 
magnitude  of  this  support- 
ing force,  the  magnitude 
and  direction  of  the  other 
supporting  force,  and  the 
stresses  in  the  members. 

SOLUTION. — Join  A 
with  D,  the  point  of  inter-  FIG.  72. 

section  of  the  line  of  direction  of  the  load  and  the  line  BE. 
Then  will  DA  be  the  direction  of  the  other  supporting  force ; 
for  the  three  external  forces,  in  order  to  form  a  balanced  sys- 
tem, must  meet  in  a  point,  except  when  they  are  parallel. 
Then  draw  ab  to  scale,  parallel  to  CD  and  equal  to  W.  From 


1 42 


APPLIED   MECHANICS. 


a  draw  ac  parallel  to  BD,  and  from  b  draw  be  parallel  to  AD  ; 
then  will  the  triangle  abca  be  the  triangle  of  external  forces, 
the  sides  ab,  be,  and  cat  taken  in  order,  representing  respectively 
the  load  W,  the  supporting  force  at  A,  and  the  supporting  force 
at  B. 

Then  from  a  draw,  ad  parallel  to  BC,  and  from  c  draw  cd 
parallel  to  AB ;  then  will  the  triangle  acd  be  the  triangle  of 
forces  for  the  joint  B,  and  the  sides  ca,  ad,  and  dc,  taken  in 
order,  will  represent  respectively  the  supporting  force  at  B,  the 
force  exerted  by  the  bar  BC  at  the  point  B,  and  the  force 
exerted  by  the  bar  AB  at  the  point  B. 

Since,  therefore,  the  force  ad  exerted  by  the  bar  CB  at  B 
is  directed  away  from  the  bar,  it  follows  that  CB  is  in  compres- 
sion ;  and,  since  the  force  dc  exerted  by  the  bar  AB  at  B  is 
directed  towards  the  bar,  it  follows  that  AB  is  in  tension. 

In  the  same  way  bdc  is  the  triangle  of  forces  for  the  point 
A ;  the  sides  be,  cdy  and  db  representing  respectively  the  sup- 
porting force  at  A,  the  force 
exerted  by  the  bar  AB  at  A, 
and  the  force  exerted  by  the 
bar  AC  at  A. 

The  bar  AB  is  again  seen  to 
be  in  tension,  as   the  force  cd 
exerted  by  the  bar  AB  at  A  is 
»d   directed  towards  the  bar. 

So  likewise  the  triangle  abd 
is  the  triangle  of  forces  for  the 
point  C. 

Fig.  73  shows  the  case  when 
the  supporting  forces  meet  the  load-line  above,  instead  of 
below,  the  truss. 

§  122.  Triangular  Frame  with  Load  and  Supporting 
Forces  Vertical.  —  Fig.  74  shows  the  construction  when  the 
load  and  also  the  supporting  forces  are  vertical.  In  this  case 


\ 


w 


FIG. 


BOW'S  NOTATION. 


143 


FIG.  74. 


the  diagram  becomes  very  much  simplified,  the  triangle  of 
external  forces  abd  becom- 
ing a  straight  line.  The 
diagram  is  otherwise  con- 
structed just  like  the  last 
one. 

§  123.   Bow's  Notation. 
—  The  notation  devised  by 
Robert  H.  Bow  very  much 
simplifies   the   construction   of    the   stress   diagrams   of   roof- 
trusses. 

This  notation  is  as  follows  :  Let  the  radiating  lines  (Fig.  75) 
represent  the  lines  of  action  of  a  system  of  forces  in  equilib- 
rium, and  let  the  polygon  abcdefa  be  the  polygon  representing 

these  forces  in  magnitude 
and  direction ;  then  denote 
the  sides  of  the  polygon 
in  the  ordinary  way,  by 
placing  small  letters  at  the 
vertices,  but  denote  the 
radiating  lines  by  capital 
letters  placed  in  the  angles. 
Thus  the  line  AB  is  the 
line  of  direction  of  the 
force  ab,  etc.  In  applying  the  notation  to  roof-trusses,  we  letter 
the  truss  with  capital  letters  in  the  spaces,  and  the  stress  dia- 
gram with  small  letters  at  the  vertices.  If,  then,  in  drawing 
the  polygon  of  equilibrium  for  any  one  joint  of  the  truss,  we 
take  the  forces  always  in  the  same  order,  proceeding  always 
in  right-handed  or  always  in  left-handed  rotation,  we  shall  be 
led  to  the  simplest  diagrams.  Hereafter  this  notation  will  be 
used  exclusively  in  determining  the  stresses  in  roof-trusses. 

§124.  Isosceles  Triangular  Frame:  Concentrated  Load 
(Fig.  76.)  —  Let  the  load  W  act  at  the  apex,  the  supporting 


FIG.  75- 


144 


APPLIED   MECHANICS. 


FIG.  76. 


forces  being  vertical ;   each  will  be  equal  to  \  W :   hence  the 
polygon  of  external  forces  will  be  the  triangle  abc,  the  sides  of 

which,  ab,  be,  and  ca,  all  lie  in 
one  straight  line.  Then  begin 
at  the  left-hand  support,  and 
proceed  again  in  right-handed 
rotation,  and  we  have  as  the  tri- 
angle of  forces  at  this  joint  cad, 
the  forces  cat  ad,  and  dc,  these 
being  respectively  the  support- 
ing force,  the  stress  in  AD,  and 
that  in  DC ;  the  directions  of 
these  forces  being  indicated  by 
the  order  in  which  the  letters  follow  each  other :  thus,  ca  is  an 
upward  force,  ad  is  a  downward  force  ;  and,  this  being  the 
force  exerted  by  the  bar  AD  at  the  left-hand  support,  we  con- 
clude that  the  bar  AD  is  in  compression.  Again  :  dc  is 
directed  towards  the  right,  or  towards  the  bar  itself,  and  hence 
the  bar  DC  is  in  tension.  The  triangle  of  forces  for  the  other 
support  is  bed,  and  that  for  the  apex  abd. 

§  125.  Isosceles  Triangular  Frame:  Distributed  Load. — 
Let  the  load  W  be  uniformly  distributed  over  the  two  rafters 
AF  and  FB  (Fig.  77)  ;  then  will 
these  two  rafters  be  subjected  to 
a  direct  stress,  and  also  to  a  bend- 
ing action  :  and  if  we  resolve  the 
load  on  each  rafter  into  two  com- 
ponents at  the  ends  of  the  rafter, 
then,  considering  these  components 
as  the  loads  at  the  joints,  we  shall 
determine  correctly  by  our  diagram  the  direct  stresses  in  all 
the  bars  of  the  truss. 

The  load  distributed  over  AF  is  — ;  and  of  this,  one-half  is 


FIG.  77. 


POLYGONAL   FRAME. 


145 


the  component  at  the  support,  and  one-half  at  the  apex,  and 
similarly  for  the  other  rafter.  This  gives  as  our  loads,  -  -  at 

each  support,  and  —  at  the  apex.     The  polygon  of  external 

2 

forces  is  eabcde,  where  the  sides  are  as  follows  :  — 

W  W     ,         W        ,       W      ,        W 

ea  =  — ,     ab  =  — ,     be  —  — ,     cd  —  — ,     de  =  — . 
42422 

Then,  beginning  at  the  left-hand  support,  we  shall  have  for  the 
polygon  of  forces  the  quadrilateral  deafd,  where  de  =  —  =  sup- 
porting force,  ea  —  --  =:  downward  load  at  support,  af  — 

4 

stress  in  AF  (compression),  fd  :=  stress  in  FD  (tension).  The 
polygon  for  the  apex  is  abf,  and  that  for  the  right-hand  support 
cdfbc. 

§  126.  Polygonal  Frame.  —  Given  a  polygonal  frame  (Fig. 
78)  formed  of  bars  jointed  together  at  the  vertices  of  the  angles, 
and  free  to  turn  on  these  joints, 
it  is  evident,  that,  in  order  that 
the  frame  may  retain  its  form, 
it  is  necessary  that  the  direc- 
tions of,  and  the  proportions 
between,  the  loads  at  the  dif- 
ferent joints,  should  bt  speci- 
ally adapted  to  the  given  form  : 
otherwise  the  frame  will  change 
its  form.  We  will  proceed  to 
solve  the  following  problem  : 
Given  the  form  of  the  frame, 
the  magnitude  of  one  load  as  AB,  and  the  direction  of  all  the 
external  forces  (loads  and  supporting  forces)  except  one,  we 
shall  have  sufficient  data  to  determine  the  magnitudes  of  all, 


FIG.  78. 


146 


APPLIED   MECHANICS. 


and  the  direction  of  the  remaining  external  forces,  and  also  the 
stresses  in  the  bars. 

Let  the  direction  of  all  the  loads  be  given,  and  also  that  of 
the  supporting  force  EF,  that  of  the  supporting  force  AF  being 
thus  far  unknown ;  and  let  the  magnitude  of  AB  be  given. 
Then,  beginning  at  the  joint  ABG,  we  have  for  triangle  of 
forces  abg  formed  by  drawing  ab  ||  and  —  AB,  then  drawing 
ga  ||  AG,  and  bg  ||  BG ;  ga  and  bg  both  being  thrusts.  Then, 
passing  to  the  joint  BCG,  we  have  the  thrust  in  BG  already 
determined,  and  it  will  in  this  case  be  represented  by  gb.  If, 
now,  we  draw  be  ||  BC,  and  gc  ||  GC,  we  shall  have  determined 
the  load  BC  as  be,  and  we  shall  have  eg  and  gb  as  the  thrusts 
in  CG  and  GB  respectively.  Continuing  in  the  same  way,  we 
obtain  the  triangles  gcd,  gde,  and  gfe,  thus  determining  the 
magnitudes  of  the  loads  cd,  de,  and  of  the  supporting  force  ef ; 
and  then  the  triangle  gaf,  formed  by  joining  a  and/,  gives  us  af 
for  the  magnitude  and  direction  of  the  left-hand  support.  The 
polygon  abcdefa  of  external  forces  is  called  the  Force  Polygon, 
while  the  frame  itself  is  called  the  Equilibrium  Polygon. 

§  127.    Polygonal    Frame    with    Loads    and    Supporting 

Forces   Vertical In  this  case  (Fig.   79)  we  may  give  the 

form  of  the  frame  and  the  mag- 
nitude of  one  of  the  loads,  to 
determine  the  other  loads  and 
the  supporting  forces,  and  also 
the  stresses  in  the  bars  ;  or  we 
may  give  the  form  of  the  frame 
and  the  magnitude  of  the  re- 
sultant of  the  loads,  to  find  the 
loads  and  supporting  forces.  In 
the  former  case  let  the  load  AB 
be  given.  Then,  proceeding  in 
the  same  way  as  before,  we  find  the  diagram  of  Fig.  79  ;  the 
polygon  of  external  forces  abcdefa  falling  all  in  one  straight  line. 


FIG.  79. 


FUNICULAR  POLYGON. —  TRIANGULAR    TRUSS. 


147 


If,  on  the  other  hand,  the  whole  load  ae  be  ^iven,  we  observe 
that  this  is  borne  by  the  stresses  in  the  extreme  bars  AG  and 
GE ;  hence,  drawing  ag  ||  AG,  and  eg  ||  EG,  we  find  eg  and  ga 
as  the  stresses  in  EG  and  GA  respectively.  Then,  proceeding 
to  the  joint  ABG,  we  find,  since  ga  is  the  force  exerted  by 
GA  at  this  point,  that,  drawing  gb  ||  GB,  we  shall  have  ab  as 
the  part  of  the  load  acting  at  the  joint  ABG,  etc. 

§  128.  Funicular  Polygon.  —  If  the  frame  of  Fig.  79  be 
inverted,  we  shall  have  the 
case  of  Fig.  80,  where  all 
the  bars,  except  FG,  are  sub- 
jected to  tension ;  FG  itself 
being  subjected  to  compres- 
sion. The  construction  of  the 
diagram  of  stresses  being  en-  \ 
tirely  similar  to  that  already  ^ 
explained  for  Fig.  79,  the  ex- 
planation will  not  be  repeated 
here.  If  the  compression 
piece  be  omitted,  the  case 
becomes  that  of  a  chain  hung 
at  the  upper  joints  (the  supporting  forces  then  becoming  iden- 
tical with  the  tensions  in  the  two  extreme  bars),  the  line  gf 
would  then  be  omitted  from  the  diagram,  and  the  polygon  of 
external  forces  would  become  abcdega. 

§  129.  Triangular  Truss  :  Wind  Pressure.  —  Inasmuch  as 
the  pressure  of  the  wind  on  a  roof  has  been  shown  by  experi- 
ment to  be  normal  to  the  roof  on  the  side  from  which  it  blows, 
we  will  next  consider  the  case  of  a  triangular  truss  with  the 
load  distributed  over  one  rafter  only,  and  normal  to  the  rafter. 

There  may  be  three  cases  :  — 

i°.  When  there  is  a  roller  under  one  end,  and  the  wind 
blows  from  the  other  side. 


FIG.  80. 


148 


APPLIED   MECHANICS. 


2°.  When  there  is  a  roller  under  one  end,  and  the  wind 
blows  from  the  side  of  the  roller. 

3°.  When  there  is  no  roller  under  either  end. 

The  last  arrangement  should  always  be  avoided  except  in 
small  and  unimportant  constructions ;  for  the  presence  of  a 
roller  under  one  end  is  necessary  to  allow  the  truss  to  change 
its  length  with  the  changes  of  temperature,  and  to  prevent  the 
stresses  that  would  occur  if  it  were  confined. 


CASE   I.  —  Using   Bow's  notation,  we  have   (Fig.  81)  the 

whole  load  represented 
in  the  diagram  by  db. 
Its  resultant  acts  at  the 
middle  of  the  rafter 
AE,  whereas  the  sup- 
porting force  at  the 
right-hand^  end  is  (in 
consequence  of  the  pres- 
ence of  the  roller)  verti- 
cal. Hence,  to  find  the 
line  of  action  of  the  other 
supporting  force,  pro- 
duce the  line  of  action 
of  the  load  till  it  meets 
a  vertical  line  drawn 

through  the  roller,  and  join  their  point  of  intersection  with  the 

support  where  there  is  no  roller.     We  thus  obtain  CD  as  the 

line  of  action  of  the  left-hand  support. 

We  can  now  determine  the  magnitude  of  the  supporting 

forces  be  and  cd  by  constructing  the  triangle  bcdb  of  external 

forces. 

Now  resolve  the  normal  distributed  force  db  into  two  single 

forces  (equal  to  each  other  in  this  case),  da  and  ab  respectively, 

acting  at  the  left-hand  support  and  at  the  apex. 


^ 


FIG.  81. 


TRIANGULAR    TRUSS:     WIND   PRESSURE. 


149 


Now  proceed  to  the  left-hand  support.  We  find  four  forces 
in  equilibrium,  of  which  two  are  entirely  known ;  viz.,  cd  and 
da:  hence,  constructing  the  quadrilateral  cdaec,  we  have  ae  as 
the  thrust  in  AE,  and  ec  as  the  tension  in  EC. 

Next  proceed  to  the  apex,  and  construct  the  triangle  of 
equilibrium  abea,  and  we  obtain  be  as  the  thrust  in  BE. 

The  triangle  bceb  is  then  the  triangle  of  equilibrium  for  the 
right-hand  sup- 
port. 

CASE  II.  - 
In  this  case 
(Fig.  82)  we  fol- 
low the  same 
method  of  pro- 
cedure, only  the 
point  of  inter- 
section of  the 
load  and  sup- 
porting forces 
is  above,  instead  of  below,  the  truss.  The  figure  explains  itself 

so  fully  that  it  is  unnecessary  to 

explain  it  here. 

CASE  III.  —  In  this  case  the 
supports  are  capable  of  exerting 
resistance  in  any  direction  what- 
ever ;  so  that,  if  any  circumstance 
should  determine  the  direction 
of  one  of  them,  that  of  the  other 
would  be  determined  also.  When  there  is  no  such  circum- 
stance, it  is  customary  to  assume  them  parallel  to  the  load 
(Fig.  83).  Making  this  assumption,  we  begin  by  dividing  the 
line  dbt  which  represents  the  load,  into  two  parts,  inversely 


FIG.  82. 


FIG.  83. 


I5O  APPLIED   MECHANICS. 

proportional  to  the  two  segments  into  which  the  line  of  action 
of  the  resultant  of  the  load  (the  dotted  line  in  the  figure) 
divides  the  line  EC.  We  thus  obtain  the  supporting  forces  be 
and  cd,  and  bcdb  is  the  triangle  of  external  forces.  We  then 
follow  the  same  method  as  in  the  preceding  cases. 

§  130.  General  Determination  of  the  Stresses  in  Roof- 
Trusses.  —  In  order  to  compute  the  stresses  in  the  different 
members  of  a  roof-truss,  it  is  necessary  first  to  know  the 
amount  and  distribution  of  the  load. 

This  consists  generally  of  — 

i°.  The  weight  of  the  truss  itself. 

2°.  The  weight  of  the  purlins,  jack-rafters,  and  superin- 
cumbent roofing,  as  the  planks,  slate,  shingles,  felt,  etc. 

3°.  The  weight  of  the  snow. 

4°.  The  weight  of  the  ceiling  of  the  room  immediately 
below  if  this  is  hung  from  the  truss,  or  the  weight  of  the 
floor  of  the  loft,  and  its  load,  if  it  be  used  as  a  room. 

5°.  The  pressure  of  the  wind ;  and  this  may  blow  from 
either  side. 

6°.  Any  accidental  load  depending  on  the  purposes  for  which 
the  building  is  used.  As  an  instance,  we  might  have  the  case 
where  a  system  of  pulleys,  by  means  of  which  heavy  weights 
are  lifted,  is  attached  to  the  roof. 

In  regard  to  the  first  two  items,  and  the  fourth,  whenever 
the  construction  is  of  importance,  the  actual  weights  should 
be  determined  and  used.  In  so  doing,  we  can  first  make  an 
approximate  computation  of  the  weight  of  the  truss,  and  use  it 
in  the  computation  of  the  stresses  ;  the  weights  of  the  ceiling 
or  of  the  floor  below  being  accurately  determined.  After  the 
stresses  in  the  different  members  have  been  ascertained  by  the 
use  of  these  loads,  and  the  necessary  dimensions  of  the  mem- 
bers determined,  we  should  compute  the  actual  weight  of  the 
truss ;  and  if  our  approximate  value  is  sufficiently  different 
from  the  true  value  to  warrant  it,  we  should  compute  again 


ST££SS£S  IN  ROOF-TRUSSES. 


the  stresses.    This  second  computation  will,  however,  seldom  be 
necessary. 

In  making  these  computations,  the  weights  of  a  cubic  foot 
of  the  materials  used  will  be  needed ;  and  average  values  are 
given  in  the  following  table  with  sufficient  accuracy  for  the 
purpose. 


WEIGHT  OF  SOME  BUILDING  MA- 
TERIALS PER  CUBIC  FOOT. 

Pounds. 

WEIGHT  OF  SLATING  PER  SQUARE 
FOOT. 
According  to  Trautwine. 

Pounds. 

TIMBER. 
Chestnut     

41 

i  inch  thick  on  laths      .     .     . 
L    «        «      «  i-inch  boards  . 

4-75 
6.75 

2C 

i.    «        «      «  ji.  «        « 

7-30 

Maple     

41 

-ft  "        "      "  laths      .    .    . 

7.00 

Oak,  live     

CQ 

T3^  "        "      "  i-inch  boards, 

9-OO 

Oak,  white  

AQ 

3       «              <«           «     jl     «              « 

9-55 

2  C  to  "}O 

i    "        "      "  laths      .     .    . 

9-25 

Pine,  yellow,  Southern      .    . 

45 

2  e  to  ^O 

i    "        "      "  i-inch  boards, 

J.     <«           «<         «    jJL   «           <« 

11.25 
11.80 

IRON. 

With  slating-felt  add      ... 
With  ^-inch  mortar  add      .     . 

ilb. 
3lbs. 

Iron  cast    ....... 

4  co 

480 

NUMBER  OF  NAILS  IN  ONE  POUND. 

No. 

Steel  

490 

J-oennv 

4  CQ 

OTHER  SUBSTANCES. 

4      " 

74O 

Asphaltum       ...              . 

80  to  90 

6      "       

I  C,O 

Mortar,  hardened     .     .          . 

lO'* 

8      "       

IOO 

Snow  freshly  fallen 

5  to   12 

I0     "       

60 

Snow,  compacted  by  rain  . 

I  c  to    co 

12      «       

40 

Slate  

140  to  1  80 

20      "                                       . 

2C 

As  to  the  weight  of  the  snow  upon  the  roof,  Stoney  recom- 
mends the  use  of  20  pounds  per  square  foot  in  moderate 
climates  ;  and  this  would  seem  to  the  writer  to  be  borne  out  by 
the  experiments  of  Trautwine  as  recorded  in  his  handbook, 


I52  APPLIED   MECHANICS. 

although  Trautwine  himself  considers  12  pounds  per  square 
foot  as  sufficient. 

§  131.  Wind  Pressure. — While  a  great  deal  of  work  has 
been  done  to  ascertain  the  direction  and  the  greatest  intensity 
of  the  pressure  of  the  wind  upon  exposed  surfaces,  as  those  of 
roofs  and  bridges,  nevertheless  the  amount  of  information  on 
the  subject  is  very  small,  inasmuch  as  but  few  experiments 
have  been  under  the  conditions  of  practice.  Before  giving  a 
summary  of  what  has  been  done  the  following  statements  will 
be  made : 

i°.  The  pressure  of  the  wind  upon  a  roof,  or  other  surface, 
is  assumed  to  be  normal  to  the  surface  upon  which  it  blows ; 
and  what  little  experimenting  has  been  done  upon  the  subject 
tends  to  confirm  this  view. 

2°.  Inasmuch  as  more  attempts  have  been  made  to  deter- 
mine experimentally  the  velocity  of  the  wind  than  its  pressure, 
hence  there  have  been  a  good  many  experiments  to  determine 
the  relation  between  the  velocity  and  the  pressure  upon  a  sur- 
face to  which  the  direction  of  the  wind  is  normal. 

3°.  A  few  experimenters  have  tried  to  determine  the  rela- 
tion between  the  intensity  of  the  pressure  on  a  surface  normal 
to  the  direction  of  the  wind  and  one  inclined  to  its  direction. 

4°.  While  the  above  have  been  the  investigations  most  com- 
monly pursued,  other  subjects  of  experiment  have  been— 

(a)  The  variation  of  pressure  with  density ;  (b)  with  tem- 
perature;  (c)  with  humidity;  (d)  with  the  size  of  surface 
pressed  upon  ;  (e)  with  the  shape  of  surface  pressed  upon ; 
(/)  whether  the  pressure  corresponding  to  a  certain  velocity  is 
the  same  whether  the  air  moves  against  a  body  at  rest,  or 
whether  the  body  moves  in  quiet  air. 

By  way  of  references  to  the  literature  of  the  subject  may 
be  given  the  following,  as  most  of  the  work  that  has  been 
done  is  included  in  them  or  in  other  references  which  they 
contain  : 


WIND  PRESSURE,  153 


i°.  Proceedings  of  the  British  Institution  of  Civil    Engineers,  vol. 
Ixix.,  year  1882,  pages  80  to  218  inclusive. 
R.  Wolff :  Treatise  on  Windmills. 

Shaler  Smith  :   Proceedings  American  Society  of  Civil  En- 
gineers, vol.  x.,  page  139. 

4°.  A.  L.  Rotch  :  Report  of  Work  of  the  Blue  Hill  Meteorological 
Observatory,  1887. 

5°.   Engineering,  Feb.  28th,  1890  :   Experiments  of  Baker. 

6°.   Engineering,   May  30,  June   6,  June   13,  1890:   Experiments   of 
O.  T.  Crosby. 

The  first  gives  an  account  of  a  very  full  discussion  of  the 
subject,  by  a  large  number  of  Engineers.  The  second  con- 
tains a  recommendation  that  the  temperature  of  the  air  be  con- 
sidered in  estimating  the  pressure.  The  fifth  gives  an  account 
of  Baker's  experiments  on  wind  pressure  in  connection  with  the 
building  of  the  Forth  Bridge. 

Before  an  account  is  given  of  the  experimental  work  that 
has  been  done,  the  following  statements  will  be  made  of  what 
are  some  of  the  methods  in  most  common  use : 

i°.  A  great  many  engineers  very  commonly  call  from  40  to 
55  pounds  per  square  foot  the  maximum  pressure  on' a  vertical 
surface  at  right  angles  to  the  direction  of  the  wind.  One  rather 
common  practice,  in  the  case  of  bridges,  is  to  estimate  30 
pounds  per  square  foot  on  the  loaded,  or  50  pounds  per  square 
foot  on  the  unloaded  structure.  Nevertheless  pressures  of  80 
and  90  pounds  per  square  foot  have  been  registered  and  re- 
corded by  the  use  of  small  pressure-plates,  and  by  computation 
from  anemometer  records. 

2°.  By  way  of  determining  the  intensity  of  the  pressure  on 
an  inclined  surface  in  terms  of  that  on  a  surface  normal  to  the 
direction  of  the  wind,  four  methods  more  or  less  used  will  be 
enumerated  here  : 

(a)  Duchemin's  formula,  which  Professor  W.  C.  Unwin 
recommends,  is  as  follows,  viz. : 


154 


A  PPLIE  D   ME  CHA  NICS. 


2  sm 


K'i  -+-  sin'  8T 

where/  =  intensity  of  normal  pressure  on  roof,  /,  =  intensity 
of  piessure  on  a  plane  normal  to  the  direction  of  the  wind, 
(b)  Hutton's  formula, 


=/,  (sin 


-84  cos  «- 


Unwin  claims  that  this  and  Duchemin's  formula  give  nearly 
the  'same  results  for  all  angles  of  inclination  greater  than  15°. 

The  following  table  gives  the  results  obtained  by  the  use  of 
each,  on  the  assumption  that  pl  =  40: 


0 

Duchemin. 

Hutton. 

1 

Duchemin. 

Hutton. 

5° 

6.89 

5-io 

50° 

38.64 

38.10 

10° 

13-59 

9.60 

55° 

39.21 

39-40 

15° 

19.32 

14.20 

60° 

39-74 

40.00 

20° 

24.24 

18.40 

65° 

39-82 

40.00 

25° 

28.77 

22.60 

70° 

39-91 

40.00 

30° 

32.00 

26.50 

75° 

39  -96 

40.00 

35° 

34-52 

30.10 

80° 

40.00 

40.00 

40° 

36.40 

33-30 

85° 

40.00 

40.00 

45° 

37-73 

36.00 

90° 

40.00 

40.00 

(c)  A  formula  very  commonly  favored,  but  which  does  not 
agree  with  any  experiments  that  have  been  made,  is 

/  =  pl  sin2  6. 

It  gives  much  lower  results,  as  a  rule,  than  either  of  the  others, 
but  it  is  favored  by  many  because,  if  we  assume  the  wind  to 
blow  in  parallel  lines  till  it  strikes  the  surface,  and  then  to  get 
suddenly  out  of  the  way,  forming  no  eddies  on  the  back  side 
of  the  surface  and  meeting  no  lateral  resistance  on  the  front 


WIND  PRESSURE.  I  55 


side,  all  of  which  are  conditions  that  do  not  exist,  we  could 
then  deduce  it  as  follows: 

Assume  a  unit  surface  making  an  angle  0  with  the  direction 
of  the  wind,  the  total  pressure  on  this  surface  in  the  direction 
of  the  wind  would  be/j  sin  0 ;  and  by  resolving  this  into  nor- 
mal and  tangential  components  we  should  have,  for  the  former, 

p  =  pi  sin2  6. 

(d)  Another  rule  which  is  sometimes  used,  but  which  has 
nothing  to  recommend  it,  is  to  consider  the  normal  intensity 
of  the  wind  pressure  per  square  foot  of  roof  surface  as  equal  to 
the  number  of  degrees  of  inclination  of  the  roof  to  the  hori- 
zontal. The  wind  pressure  allowed  for  by  this  rule  is  very 
excessive,  as  it  would  be  90  pounds  per  square  foot  for  a  ver- 
tical surface. 

Taking  up,  now,  the  experimental  work  that  has  been  done, 
we  will  begin  with  the  attempts  to  determine  velocities  and 
pressures,  and  the  relation  between  them. 

i°.  In  regard  to  velocities,  these  are  determined  by  using 
some  kind  of  an  anemometer,  and  in  all  these  cases  there  are 
several  difficulties  and  sources  of  error,  as  follows : 

(a)  In  many  cases  the  anemometers  have  not  even  been 
graduated  experimentally,  but  it  has  been  assumed  outright 
that  the  velocity  of  the  air  is  just  three  times  the  linear  velocity 
of  the  cups  of  a  cup  anemometer. 

(&)  When  they  have  been  graduated,  it  has  generally  been 
done  by  attaching  them  to  the  end  of  the  arm  of  a  whirling 
machine,  which,  when  the  arm  is  long,  and  the  velocity  moder- 
ate, will  do  very  well,  but  is  the  more  inaccurate  the  shorter 
the  arm  and  the  higher  the  velocity  of  motion. 

(c)  The  wind  always  comes  in  gusts,  and  hence  the  ane- 
mometer does  not  register  the  average  velocity  of  the  wind  at 
any  one  moment,  but  that  of  the  particular  portion  that  comes 


156  APPLIED  MECHANICS. 

in  contact  with  it,  and  this  is  always  a  small  portion,  on  ac- 
count of  the  small  size  of  the  anemometer. 

(d)  In  order  to  get  an  indication  which  is  not  affected  by 
the  cross-currents  reflected  from  the  surrounding  buildings  and 
chimneys,  it  is  necessary  to  put  the  anemometer  very  high  up, 
and  then,  of  course,  we  obtain  the  indications  corresponding 
to  that  height,  which  is  greater  than  that  of  the  buildings,  and 
it  is  well  known  that  the  velocity  of  the  wind  increases  very 
considerably  with  the  height. 

Next,  as  to  the  direct  determination  of  pressure,  this  has 
usually  been  done  by  means  of  some  kind  of  pressure-plate, 
either  round  or  square,  but  of  small  size,  thus  allowing  the 
eddies  formed  on  the  back  side  of  the  plate  to  have  a  con- 
siderable effect.  The  results  obtained  by  the  use  of  different 
sizes  and  different  shapes  of  plates  have  therefore  differed  very 
considerably ;  and  while  some  have  claimed  that  the  pressure 
per  square  foot  increases  with  the  size  of  the  surface  pressed 
upon,  it  has  been  very  thoroughly  proved  by  the  more  modern 
investigations  that  the  opposite  is  true,  and  that  the  pressure 
decreases  with  the  size. 

While  the  records  from  small  pressure-plates  have  fre- 
quently shown  very  high  pressures  per  square  foot,  as  80,  90, 
or  even  over  100  pounds  per  square  foot,  it  has  become  very 
generally  recognized  by  engineers  that  by  far  the  greater  part 
of  existing  buildings  and  bridges  would  be  overturned  by  winds 
of  such  force,  or  anywhere  near  such  force,  and  it  has  not  been 
customary  among  them  to  make  use  of  such  high  figures  for 
wind  pressure  on  bridges  and  roofs  in  computing  the  stability 
of  structures.  While  some  of  the  figures  in  general  use  have 
already  been  given,  nevertheless  the  tendency  of  modern  inves- 
tigation seems  to  be  to  obtain  rather  lower  figures.  In  this  con- 
nection it  is  well  to  refer  to  the  work  done  by  Baker  in  connec- 
tion with  the  construction  of  the  Forth  Bridge.  The  following 
description  is  taken  from  "  Engineering"  of  Feb.  28th,  1890: 


WIND  PRESSURE. 


157 


"  The  wind  pressure  to  be  provided  for  in  the  calcu- 
lations for  bridges  in  exposed  positions  is  56  Ibs.  per  square 
foot,  according  to  the  Board  of  Trade  regulations,  and  this 
twice  over  the  whole  area  of  the  girder  surface  exposed,  the 
resistance  to  such  pressure  to  be  by  dead-weight  in  the  struc- 
ture alone. 

"  The  most  violent  gales  which  have  occurred  during  the 
construction  of  the  Forth  Bridge  are  given,  with  the  pressures 
recorded  on  the  wind  gauges,  in  the  annexed  table : 


Year. 

Month 
and 
Day. 

Pressure  in  pounds  per  square  foot. 

Direction 
of 
Wind. 

Revolving 
Gauge. 

Small 
fixed 
Gauge. 

Large 
fixed 
Gauge. 

In  centre 
of  large 
Gauge. 

Right- 
hand  top 
of  large 
Gauge. 

1883 

Dec.    II, 

33 

39 

22 

S.  W.* 

1884 

Jan.    26, 

65 

41 

35 

S.  W.* 

1884 

Oct.    27, 

29 

23 

18 

S.  W. 

1884 

Oct.    28, 

26 

29 

19 

S.  W. 

1885 

Mar.  20, 

30 

25 

I? 

W. 

1885 

Dec.     4, 

25 

27 

J9 

W. 

1886 

Mar.  31, 

26 

31 

19 

S.  W. 

1887 

Feb.     4, 

26 

41 

15 

S.  W. 

1888 

Jan.      5, 

27 

16 

7 

S    E. 

1888 

Nov.  17, 

35 

4i 

27 

W. 

1889 

"          2, 

27 

34 

12 

S.  W. 

1890 

Jan.    19, 

27 

28 

16 

S.  VV. 

1890 

"         21, 

26 

3S 

T5 

W. 

1890 

"         22, 

27 

24 

18 

23i 

22 

S.  W.  by  W. 

*  These  data  are  unreliable,  owing-  to  faulty  registration  by  the  indicator-needle  as  will 
be  presently  explained.  They  were  altered  after  this  date.  The  barometer  fell  to  27.5  inches 
on  that  occasion,  over  three  quarters  of  an  inch  within  an  hour. 


158  APPLIED   MECHANICS. 

"The  pressure-gauges,  which  were  put  up  in  the  summer 
of  1882  on  the  top  of  the  old  castle  of  Inchgarvie,  and  from 
which  daily  records  have  been  taken  throughout,  were  of  very 
simple  construction.  The  maximum  pressures  only  were  taken. 
The  most  unfavorable  direction  from  which  the  wind  pressure 
can  strike  the  bridge  is  nearly  due  east  and  west,  and  two  out 
of  the  three  gauges  were  fixed  to  face  these  directions,  while 
a  third  was  so  arranged  as  to  register  for  any  direction  of 
wind. 

"The  principal  gauge  is  a  large  board  20  feet  long  by  15 
feet  high,  or  300  square  feet  area,  set  vertically  with  its  faces  east 
and  west.  The  weight  of  this  board  is  carried  by  two  rods  sus- 
pended from  a  framework  surrounding  the  board,  and  so  ar- 
ranged as  to  offer  as  little  resistance  as  possible  to  the  passage 
of  the  wind,  in  order  not  to  create  eddies  near  the  edge  of  the 
board.  In  the  horizontal  central  axis  of  the  board  there  are 
fixed  two  pins  which  fit  into  the  lower  eyes  of  the  suspension- 
rods,  the  object  being  to  balance  the  board  as  nearly  as  pos- 
sible. Each  of  the  four  corners  of  the  board  is  held  between 
two  spiral  springs,  all  carefully  and  easily  adjusted  so  that  any 
pressure  exerted  on  either  face  will  push  it  evenly  in  the  op- 
posite direction,  but  upon  such  pressure  being  removed  the 
compressed  springs  will  force  the  board  back  to  its  normal 
position.  To  the  four  corners  four  irons  are  attached,  uniting 
in  a  pyramidal  formation  in  one  point,  whence  a  single  wire 
passes  over  a  pulley  to  the  registering  apparatus  below.  In 
order  to  ascertain  to  some  extent  how  far  great  gusts  of  wind 
are  quite  local  in  their  action,  and  exert  great  pressure  only 
upon  a  very  limited  area,  two  circular  spaces,  one  in  the  exact 
centre  and  one  in  the  right-hand  top  corner,  about  18  inches 
in  diameter,  were  cut  out  of  the  board  and  circular  plates  in- 
serted, which  could  register  independently  the  force  of  the 
wind  upon  them. 

"By  the  side  of  the  large  square  board,  at  a  distance  of 


WIND   PRESSURE.  159 


about  8  feet,  another  gauge,  a  circular  plate  of  1.5  square  feet 
area,  facing  east  and  west,  was  fixed  up  with  separate  regis- 
tration. This  was  intended  as  a  check  upon  the  indications 
given  by  the  large  board. 

"Another  gauge  of  the  same  dimensions  as  the  last,  but 
with  the  disc  attached  to  the  short  arm  of  a  double  vane,  so 
that  it  would  face  the  wind  from  whatever  direction  it  might 
come,  was  set  up. 

"  On  one  occasion  the  small  fixed  board  appeared  to  regis- 
ter 65  pounds  to  the  square  foot — a  registration  which  caused 
no  little  alarm  and  anxiety.  Mr.  Baker  found,  upon  inves- 
tigation, that  the  registering  apparatus  was  not  in  good  order, 
and  after  adjusting  it  the  highest  pressure  recorded  was  41 
pounds. 

"  In  order  to  determine  the  effect  of  the  wind  upon  surfaces 
like  that  of  the  exposed  surface  of  the  bridge,  he  devised  an 
apparatus  which  consisted  of  a  light  wooden  rod  suspended  in 
the  middle,  so  as  to  balance  correctly,  by  a  string  from  the 
ceiling.  At  one  end  was  attached  a  cardboard  model  of  the 
surface,  the  resistance  of  which  was  to  be  tested,  and  at  the 
opposite  end  was  placed  a  sheet  of  cardboard  facing  the  same 
way  as  the  model,  so  arranged  that  by  means  of  another  and 
adjustable  sheet,  which  would  slide  in  and  out  of  the  first, 
the  surface  at  that  end  could  be  increased  or  decreased  at 
the  will  of  the  operator.  The  mode  of  working  is  for  a 
person  to  pull  it  from  its  perpendicular  position  towards 
himself,  and  then  gently  release  it,  being  careful  to  allow 
both  ends  to  go  together.  If  this  is  properly  done,  it  is  evi- 
dent that  the  rod  will  in  swinging  retain  a  position  parallel 
to  its  original  position,  supposing  that  the  model  at  one 
end  and  the  cardboard  frame  at  the  other  are  balanced  as 
to  weight,  and  that  the  two  surfaces  exposed  to  the  air 
pressure  coming  against  it  in  swinging  are  exactly  alike. 
Should  one  area  be  greater  than  the  other,  the  model  or  card- 


l6o  APPLIED  MECHANICS. 

board*  sheet,  whichever  it  may  be,  will  b~  lagging  behind,  and 
twist  t^e  string." 

The  experiments  carried  on  in  various  ways  by  different 
people  and  at  different  times  are  generally  in  agreement  with 
each  other  and  with  the  results  of  more  elaborate  processes. 
The  information  specially  desired  was  in  regard  to  the  wind 
pressure  upon  surfaces  more  or  less  sheltered  by  those  imme- 
diately in  front  of  them.  In  this  regard  Mr.  Baker  satisfied 
himself  that,  while  the  results  differed  very  considerably  ac- 
cording to  the  distance  apart  of  the  surfaces,  in  no  case  was 
the  area  affected  by  the  wind,  in  any  girder  which  had  two  or 
more  surfaces  exposed,  more  than  1.8  times  the  area  of  the 
surface  directly  fronting  the  wind,  and,  as  the  calculations  had 
been  made  for  twice  this  surface,  the  stresses  which  the  struc- 
ture will  receive  from  this  cause  will  be  less  than  those  pro- 
vided for. 

Next,  as  to  the  relation  between  velocity  and  pressure,  a 
great  many  formulae  have  been  devised,  to  agree  with  the 
results  of  different  experimenters.  Most  all  of  them  make  the 
pressure  proportional  to  the  square  of  the  velocity  ;  while 
some  add  a  term  proportional  to  the  velocity  itself,  and  when 
higher  velocities  are  reached,  as  those  usual  in  gunnery,  terms 
have  been  introduced  with  powers  of  the  velocity  higher  than 
the  second.  It  is  hardly  worth  while  to  consider  these  dif- 
ferent formulae,  as  it  is  rather  the  pressure  than  the  velocity 
that  the  engineer  is  interested  in,  and  correct  information  in 
this  regard  is  to  be  obtained  rather  from  pressure-boards  than 
from  anemometers.  Nevertheless,  it  may  be  stated  that  one 
of  the  most  usual  formulae  is  that  of  Smeaton,  and  is 


200 

where  P=  pressure  in  pounds  per  square  foot,  and  F=  velocity 


WIND  PRESSURE.  l6l 


in  miles  per  hour.  This  formula  agrees  very  well  with  a  num- 
ber of  experiments  that  have  been  made  where  anemometers 
have  been  used  to  determine  the  velocity,  and  small  pressure- 
plates  (say  one  square  foot)  to  determine  the  pressure  ;  thus 
this  formula  satisfies  very  well  the  experiments  made  at  the 
Blue  Hill  Meteorological  Observatory,  near  Boston,  Mass., 
U.  S.  A. 

It  was  originally  deduced  from  some  very  old  experiments 
of  Rouse  ;  and  it  agrees  with  a  good  many,  but  disagrees  with 
other  experiments.  It  is  probably  the  formula  that  has  been 
more  quoted  than  any  other. 

A  little  ought  also  to  be  said  in  regard  to  the  pressure  of 
the  wind  on  very  high  structures,  as  on  the  piers  of  high  via- 
ducts and  on  tall  chimneys.  In  this  regard  it  is  to  be  ob- 
served : 

i°.  The  pressure,  as  well  as  the  velocity  of  the  wind,  be- 
comes greater  the  higher  up  from  the  ground  the  surface  ex- 
posed is  situated. 

2°.  From  calculations  on  chimneys  that  have  stood  for  a 
long  time,  Rankine  deduced,  as  the  greatest  average  wind 
pressure  that  could  be  realized  in  the  case  of  tall  chimneys,  55 
pounds  per  square  foot. 

3°.  In  making  the  piers  of  high  viaducts,  it  would  seem 
desirable  not  to  make  them  solid,  but  to  use  only  four  up- 
rights at  the  corners  connected  by  lattice  work,  in  order  to 
expose  a  smaller  surface  to  the  wind.  Nevertheless,  as  was  ex- 
plained, it  will  not  do  to  separate  the  structure  into  its  com- 
ponent parts,  and  to  estimate  the  pressure  on  each  part 
separately  and  then  add  the  results  together  to  get  the  total 
effect ;  but  we  really  need  some  such  experiments  as  those  of 
Baker. 

4°.  Some  old  experiments  of  Borda  bear  out  the  common 
practice  of  assuming  the  wind  pressure  on  the  surface  of  a  cir- 


1 62  APPLIED  MECHANICS 

cular  cylinder  one  half  that  which  would  exist  on  its  projection 
on  a  plane  normal  to  the  direction  of  the  wind. 

There  remains  now  only  to  refer  to  a  serial  article  by  O.  T. 
Crosby,  in  "  Engineering"  of  May  30,  June  6th,  and  June  I3th, 
1890,  containining  some  experiments  made  by  him  on  wind 
pressure  near  Baltimore,  Md.  The  first  two  numbers  contain 
rather  a  summary  of  what  has  been  done  by  others,  and  it  is 
in  the  copy  of  June  I3th  that  is  to  be  found  the  account  of  his 
own  work,  which  was  done  in  order  to  determine  the  resistances 
of  the  air  to  fast-moving  trains. 

He  used  a  whirling  arrangement  turning  about  a  vertical 
axis,  to  the  end  of  which  was  attached  a  car,  the  circumference 
through  which  the  car  moved  being  36  feet. 

In  order  to  determine  whether  the  circular  motion  produced 
any  disturbing  effect,  he  ran  a  car  having  a  cross-section  of  5.1 
square  feet  on  a  circular  track  about  two  miles  in  circumference, 
the  speed  of  the  car  being  about  50  miles  per  hour,  and  the 
results  obtained  in  this  way  agreed  very  nearly  with  those  ob- 
tained from  his  whirling  table.  The  special  peculiarity  of  his 
results  is  that  he  obtained,  by  plotting  them,  the  law  that  the 
pressure  varies  directly  as  the  first  power  of  the  velocity,  and 
not  as  the  square  or  some  higher  power;  also,  his  pressures, 
after  the  velocity  had  passed  25  or  30  miles  per  hour,  are 
much  lower  than  those  given  by  Smeaton  and  others,  the  pres- 
sure on  a  normal  plane  surface  moving  at  115  miles  per  hour 
being  about  27  pounds  per  square  foot. 

The  cars  used  were  generally  about  3  feet  long  without  the 
front.  The  fronts  attached  were:  i°.  Normal  plane  surface; 
2°.  Wedge,  base  i,  height  I  ;  3°.  Pyramid,  base  I,  height  2;  4°. 
Wedge  and  cyma,  base  I,  height  2;  5°.  Parabolic  wedge, 
base  I,  height  2. 

His  experiments  covered  a  range  of  velocities  from  30  to 
130  miles  per  hour. 


DISTRIBUTION  OF   THE   LOADS.  163 

The  law  of  the  first  powers  of  the  velocities  seems  peculiar, 
and  certainly  ought  not  to  be  accepted  without  further  cor- 
roborative evidence  ;  but  the  low  values-  of  the  pressures  agree 
with  Baker's  results  and  with  the  tendency  of  the  more  modern 
investigations. 

§  132.  Approximate  Estimation  of  the  Load In  all 

important  constructions,  the  estimates  of  the  loads  should  be 
made  as  above  described.  For  smaller  constructions,  and  for 
the  purposes  of  a  preliminary  computation  in  all  cases,  we  only 
estimate  the  roof-weight  roughly ;  and  some  authors  even  as- 
sume the  wind  pressure  as  a  vertical  force. 

Trautwine  recommends  the  use  of  the  following  figures  for 
the  total  load  per  square  foot,  including  wind  and  snow,  when 
the  span  is  75  feet  or  less  :  — 

Roof  covered  with  corrugated  iron,  unbearded     .     .     .28  Ibs. 

Roof  plastered  below  the  rafters 38 .  " 

Roof,  corrugated  iron  on  boards 31  " 

Roof  plastered  below  the  rafters 41  " 

Roof,  slate,  unbearded  or  on  laths 33  " 

Roof,  slate,  on  boards  ij  inches  thick 35  " 

Roof,  slate,  if  plastered  below  the  rafters 46  " 

Roof,  shingles  on  laths    .     .     .     ,     »,-,*,.  •    .•     •     •     •  3°  " 

Roof  plastered  below  rafters  or  below  tie-beam    ...  40  " 
From  75  to  100  feet,  add  4  Ibs.  to  each. 

§  133.  Distribution  of  the  Loads.  —  The  methods  for  de- 
termining the  stresses,  which  will  be  used  here,  give  correct 
results  only  when  the  loads  are  concentrated  at  joints,  and  are 
not  distributed  over  any  members  of  the  truss. 

In  constructions  of  importance,  this  concentration  of  the 
loads  at  the  joints  should  always  be  effected  if  possible; 
because,  when  this  is  the  case,  the  stresses  in  the  members 
of  the  truss  can  be,  if  properly  fitted,  obliged  to  resist  only 
stresses  of  direct  tension,  or  of  direct  compression  ;  but,  when 
there  is  a  load  distributed  over  any  member  of  the  truss,  that 
member,  in  addition  to  the  direct  compression  or  direct  tension, 
is  subjected  to  a  bending-stress.  The  effect  of  this  bending 


i64 


APPLIED  MECHANICS. 


cannot  be  discussed  until  we  have  studied  the  theory  of  beams. 
Nevertheless,  it  is  a  fact  that  we  have  no  experimental  knowl- 
edge of  the  behavior  of  a  piece  under  combined  compression 
and  bending ;  and  we  know  that  when  certain  pieces  are  to 
resist  bending,  in  addition  to  tension,  they  must  be  made  much 
larger  in  proportion  than  is  necessary  when  they  bear  tension 
only. 


FIG.  84. 

• 

The  manner  in  which  this  concentration  of  the  loads  is 
effected,  is  shown  in  Fig.  84,  which  is  intended  to  represent  one 
of  a  series  of  trusses  that  supports  a  roof,  the  rafters  being  the 
two  lower  ones  in  the  figure.  Resting  on  two  consecutive 
trusses,  and  extending  from  one  to  the  other,  are  beams  called 
purlins,  which  should  be  placed  only  above  the  joints  of  the  truss, 
and  which  are  shown  in  cross-section  in  the  figure.  On  these 
purlins  are  supported  the  jack-rafters  parallel  to  the  rafters,  and 
at  sufficiently  frequent  intervals  to  support  suitably  the  plank 
and  superincumbent  roofing-materials. 

By  this  means  we  secure  that  the  entire  bending-stress  comes 
upon  the  jack-rafters  and  purlins,  and  that  the  rafters  proper 
are  subjected  only  to  a  direct  compression.  When,  however, 
the  load  is  distributed,  i.e.,  when  the  roofing  rests  directly  on  the 
rafters,  or  when  the  purlins  are  placed  at  points  other  than  the 
joints,  the  bending-stress  should  be  taken  into  account ;  and 
while  the  methods  to  be  developed  here  will  give  the  stresses 


DIRECT  DETERMINATION  OF   THE  STRESSES.  165 

in  all  the  members  that  are  not  subjected  to  bending,  the  ben.d- 
ing-stress  may  be  largely  in  excess  of  the  direct  stress  in  those 
pieces  that  are  subjected  to  bending.  How  to  take  this  into 
account  will  be  explained  later. 

Another  important  item  to  consider  is,  that,  in  constructions 
of  importance,  a  roller  should  be  placed  under  one  end  of  the 
truss  to  allow  it  to  move  with  the  change  of  temperature ;  as 
otherwise  some  of  the  members  will  be  either  bent,  or  at  least 
subjected  to  initial  stresses.  The  presence  of  a  roller  obliges 
the  supporting  force  at  that  point  to  be  vertical,  whether  the 
load  be  vertical  or  inclined. 

It  is  customary,  and  does  not  entail  any  appreciable  error, 
to  consider  the  weight  of  the  truss  itself,  as  well  as  that  of  the 
superincumbent  load,  as  concentrated  at  the  upper  joints ;  i.e., 
those  on  the  rafters. 

In  the  case  of  a  ceiling  on  the  room  below,  or  of  a  loft 
whose  floor  rests  on  the  lower  joints,  we  must,  of  course,  ac- 
count the  proper  load  as  resting  on  the  lower  joints. 

§134.  Direct  Determination  of  the  Stresses.  —  This,  as 
we  have  seen,  is  merely  a  question  of  equilibrium  of  forces  in 
a  plane,  where  certain  forces  acting  are  known,  and  others  are 
to  be  determined. 

As  to  the  methods  of  solution,  we  might  adopt  — 

i°.  A  graphical  solution,  laying  off  the  loads  to  scale,  and 
constructing  the  diagram  by  the  use  of  the  propositions  of 
the  polygon,  and  the  triangle  of  forces,  and  then  determining  the 
results  by  measuring  the  lines  representing  the  stresses  to 
the  same  scale. 

2°.  An  analytical  solution,  imposing  the  analytical  conditions 
of  equilibrium,  as  given  under  the  "  Composition  of  Forces," 
between  the  known  and  unknown  forces. 

3°.  A  third  method  is  to  construct  the  diagram  as  in  the 
graphical  solution,  but  then,  instead  of  determining  the  results 
by  measuring  the  resulting  lines  to  scale,  to  compute  the  un« 


1 66  APPLIED  MECHANICS. 

known  from  the  known  lines  of  the  diagram  by  the  ordinary 
methods  of  trigonometry. 

The  first,  or  purely  graphical,  method,  is  one  which  has 
received  a  very  large  amount  of  attention  of  late  years,  and 
in  which  a  great  deal  of  progress  has  been  made.  It  is,  doubt- 
less, very  convenient  for  a  skilled  draughtsman,  and  especially 
convenient  for  one  who,  though  skilled  in  draughting,  is  not 
free  with  trigonometric  work ;  but  it  seems  to  me,  that,  when 
the  results  are  determined  by  computation  from  the  diagram, 
there  is  less  chance  of  a  slight  error  in  some  unfavorable  tri- 
angle vitiating  all  the  results.  I  am  therefore  disposed  to 
recommend  for  roof-trusses  the  third  method. 

In  the  case  of  bridge-trusses,  on  the  other  hand,  I  believe 
the  graphical  not  to  be  as  convenient  as  a  purely  analytic 
method. 

§  135.  Roof-Trusses.  —  In  what  follows,  the  graphical  solu- 
tions will  be  explained  with  very  little  reference  to  the  trigono- 
metric work,  as  that  varies  in  each  special  case,  and  to  one  who 
has  a  reasonable  familiarity  with  the  solution  of  plane  triangles, 
it  will  present  no  difficulty  ;  whereas  to  deduce  the  formulae 
for  each  case  would  complicate  matters  very  much.  Proceed- 
ing to  special  examples,  let  us  take,  first,  the  truss  shown  in 
Fig.  85,  and  let  the  vertical  load  upon  it  be  W  uniformly  dis- 
tributed over  the  top  of  the  roof,  the  purlins  being  at  the  joints 
on  the  rafters. 

The  loads  at  the  several  joints  will  then  be  as  follows,  viz. 
(Fig.  85*),  — 

W  W 

ab  =  kl  =  — ,    be  =  ed  =  de  =  ef  =  fg  =  gh  =  hk  =  — . 
16  8 


Then  the  supporting  forces  will  be 

lm  =  ma  - 
We  thus  have,  as  polygon  of  external  forces,  abcdcfghklma. 


W 
lm  =  ma  =  — . 

2 


KOOF-TKUSSES. 


I67 


Now  proceed  to  either  support,  say,  the  left-hand  one ;  and 
there  we  have  the  two  forces  ab  and  ma  known,  while  by  and 
ym  are  unknown.  We  then  construct 
the  quadrilateral  maby  in  the  figure,  and 
thus  determine  by  and  ym.  As  to  whether 


\ 


"fghkl 


dbcde 


FIG.  85. 

these  represent  thrust  or  tension, 
we  need  only  remember  that  they 
are  the  forces  exerted  by  the  re- 
spective bars  at  the  joints :  and,  since  by  is  directed  away  from 
the  bar  BY,  this  bar  is  in  compression;   whereas,  ym  being 
directed  towards  the  bar  YM,  that  bar  is  in  tension. 


APPLIED   MECHANICS, 


Having  determined  these  two  stresses,  we  next  proceed  to 
another  joint,,  where  we  have  only  two  unknown  forces.  Take 
the  joint  at  which  the  load  be  acts,  and  we  have  as  known 
quantities  the  load  be,  and  also  the  force  exerted  by  the  bar 
YBy  which  is  in  compression.  This  force  is  now  directed  away 
from  the  bar,  and  hence  is  represented  by  yb.  The  unknown 
forces  are  the  stresses  in  CX  and  X  Y.  Hence  we  construct 
the  quadrilateral  cxybc  ;  and  we  thus  determine  the  stresses  in 
CX  and  X  Y  as  ex  and  xyt  both  being  thrusts. 

Next  proceed  to  the  joint  YXW,  and  construct  the  quadri- 
lateral myxwm,  and  thus  determine  the  tension  xw  and  the 
tension  wm. 

Next  proceed  to  the  joint  where  cd  acts,  and  so  on.  We 
thus  obtain  the  diagram  (Fig.  8$a)  giving  all  the  stresses. 

The  truss  in  the  figure  was  constructed  with  an  angle  of  30* 
at  the  base,  and  hence  gives  special  values  in  accordance  with 
that  angle. 

In  order  to  show  how  we  may  compute  the  stresses  from  the 
diagram,  the  computation  will  be  given. 

From  triangle  bmy,  we  have  bm  =  —  W 


ym  =  --ZTcot  30°  = 

16  16 


by  =  -^cosec  30°  =  ?-W  '  =  ky. 
16  8 


From  the  triangle  umc,  we  have  cm  =  -      W, 


um  = 

16 


ROOF-TRUSSES. 


160 


yx  =  yw  sec  30°  =  [  ^3  ^)A  =  !¥  =  w  =  vt, 
\i6      /v/3 


wv  = 


=  W 


256       256        16 


=  wm  sec  30 


sec  30°  = 


256  256 


16 


Hence  we  shall  have  for  the  stresses,  — 

VERTICALS  (tension). 


=     W- 


RAFTERS  (compression). 

by  =  kn 

ex  =  ho 

dv  —  gq 

ct  =  fs 


HORIZONTAL  TIES  (tension). 
my    =  mn 

mw  —  mp 

mu   =  mr 

16 


16 


=  op 
=  qr 


W 
16 
W_ 

8' 

I1 


DIAGONAL  BRACES  (compression) 

-  K 
8  ' 


xy  =  on 
WV  =  qp 
tu  =  sr 


16 
~8~ 


I/O  APPLIED   MECHANICS. 

Next,  as  to  the  stresses  due  to  wind  pressure,  we  will  sup- 
pose that  there  is  a  roller  under  the  left-hand  end  of  the  truss, 
and  none  under  the  right-hand  end ;  and  we  will  proceed  to 
determine  the  stresses  due  to  wind  pressure. 

First,  suppose  the  wind  to  blow  from  the  left-hand  side  of 
the  truss,  and  let  the  total  wind  pressure  be  (Fig.  85^)  af=  IV,. 
The  resultant,  of  course,  acts  along  the  dotted  line  drawn  per- 
pendicular to  the  left-hand  rafter  at  its  middle  point,  as  shown 
in  Fig.  85. 

The  left-hand  supporting  force  will  be  vertical :  hence,  pro- 
ducing the  above-described  dotted  line,  and  a  vertical  through 
the  roller  to  their  intersection,  and  joining  this  point  with  the 
right-hand  end  of  the  truss,  we  have  the  direction  of  the  right- 
hand  supporting  force.  In  this  case,  since  the  angle  of  the 
truss  is  30°,  the  line  of  action  of  the  right-hand  supporting 
force  coincides  in  direction  with  the  right-hand  rafter.  We 
now  construct  the  triangle  of  external  forces  afm,  and  we 
obtain  the  supporting  forces  fm  and  ma.  We  then  have,  as 
the  loads  at  the  joints, 


be  =  -^  =  cd  =  de. 
4 

Then  proceed  as  before  to  the  left-hand  joint ;  and  we  find  that 
two  of  the  four  forces  acting  there  are  known,  viz.,  ma  and  ab, 
and  two  are  unknown,  viz.,  the  stresses  in  .Z>  Fand  YM.  Then 
construct  the  quadrilateral  mabym,  and  we  have  the  stresses  by 
and  ym ;  the  first  being  compression  and  the  second  tension, 
as  shown  by  reasoning  similar  to  that  previously  adopted. 

Then  pass  to  the  next  joint  on  the  rafter,  and  construct  the 
quadrilateral  ybcxy,  where  yb  and  be  are  already  known,  and  we 
obtain  ex  and  xy  ;  and  so  proceed  as  before  from  joint  to  joint, 


R O OF- TR USSES    WITH  LOADS  AT  LO  WER  JOINTS.       \  7 T 

remembering,  that,  in  order  to  be  able  to  construct  the  polygon 
of  forces  in  each  case,  it  is  necessary  that  only  two  of  the  forces 
acting  should  be  unknown. 

When  the  wind  blows  from  the  other  side,  we  shall  obtain 
the  diagram  shown  in  Fig.  85^.  A 

After  having  determined  the  stresses  from  the  vertical  load 
diagram  and  those  from  the  two  wind  diagrams,  we  should,  in 
order  to  obtain  the  greatest  stress  that  can  come  on  any  one 
member  of  the  truss,  add  to  the  stress  due  to  the  vertical  load 
the  greater  of  the  stresses  due  to  the  wind  pressure. 

§  136.  Roof-Truss  with  Loads  at  Lower  Joints.  —  In 
Fig.  86  is  drawn  a  stress  diagram 
for  the  truss  shown  in  Fig.  84  on 
the  supposition  that  there  is  also  X 
a  load  on  the  lower  joints.  In 
this  case  let  W  be  the  whole  load 
of  the  truss,  except  the  ceiling, 
and  Wi  the  weight  of  the  ceiling 
below  ;  the  latter  being  supported 
at  the  lower  joints  and  on  the 
two  extreme  vertical  suspension  FlG- 86- 

rods.     Then  will  the  loads  at  the  joints  be  as  follows;  viz., — 


X    >< 


ab    = 
be    = 

cd    = 
mn 


+ 


=    -g  IV  i  —  rq 


=  hk, 

=  gh  =  de  =  fg  =  e/9 

=  on  =  qp  =  op. 


Observe  that  there  is  no  joint  at  the  lower  end  of  either  of  the 
end  suspension  rods,  but  that  whatever  load  is  supported  by 
these  is  hung  directly  from  the  upper  joints,  where  be  and  hk  act. 
We  have  also  for  each  of  the  supporting  forces  Im  and  ra 


APPLIED   MECHANICS. 


Hence  we  have,  for  the  polygon  of  external  forces, 
abcdefghklm  nopqra, 

which  is  all  in  one  straight  line,  and  which  laps  over  on 
itself. 

In  constructing  the  diagram,  we  then  proceed  in  the  same 
way  as  heretofore. 

§  137.  General  Remarks.  —  As  to  the  course  to  be  pursued 
in  general,  we  may  lay  down  the  following  directions  :  — 

I  °.  Determine  all  the  external  forces  ;  in  other  words,  the  loads 
being  known,  determine  the  supporting  forces. 

2°.  Construct  the  polygon  of  forces  for  each  joint  of  the  truss, 
beginning  at  some  joint  where  only  two  of  the  forces  acting  at 
that  joint  are  unknown.  This  is  usually  the  case  at  the  support. 
Then  proceed  from  joint  to  joint,  bearing  in  mind  that  we  can 
only  determine  the  polygon  of  forces  when  the  magnitudes  of 
all  but  two  sides  are  known. 

3°.  Adopt  a  certain  direction  of  rotation,  and  adhere  to  it 
throughout;  i.e.,  if  we  proceed  in  right-handed  rotation  at  one 
joint,  we  must  do  the  same  at  all,  and  we  shall  thus  obtain  neat 
and  convenient  figures. 

4°.  Observe  that  the  stresses  obtained  are  the  forces  exerted 
by  the  bars  tinder  consideration,  and  that  these  are  thrusts  when 
they  act  away  from  the  bars,  and  tensions  ivhen  they  are  directed 
towards  the  bars. 

We  will  next  take  some  examples  of  roof-trusses,  and  con- 
struct the  diagrams  of  some  of  them,  calling  attention  only  to 
special  peculiarities  in  those  cases  where  they  exist. 

It  will  be  assumed  that  the  student  can  make  the  trigono- 
metric computations  from  the  diagram. 

The  scale  of  load  and  wind  diagram  will  not  always  be  the 
same  ;  and  the  stress  diagrams  will  in  general  be  smaller  than 
is  advisable  in  using  them,  and  very  much  too  small  if  the 


ROOF-TRUSSES   WITH  LOADS  AT  LOWER  JOINTS.        173 

results  were  to  be  obtained  by  a  purely  graphical  process  with- 
out any  computation. 

The  loads  will  in  all  cases  be  assumed  to  be  distributed 
uniformly  over  the  jack-rafters,  or,  in  other  words,  concen- 
trated at  the  joints. 

Those  cases  where  no  stress  diagram  is  drawn  may  be  con- 
sidered as  problems  to  be  solved. 


FIG.  87. 


FIG.  87^. 


174 


APPLIED   MECHANICS. 


abode 


\p         ^Zuvwxyz^yg/ 

>/ 


FIG.  88c. 


ROOF-TRUSSES    WITH  LOADS  AT  LOWER  JOINTS.       175 


FIG.  89. 


FIG.  8ga. 


FIG.  90. 


FIG.  goa. 


FIG.  91. 


a 

y^\ 

b           - 

E 

C        A 

ts/ 

o 

"       A 

0 

K 

H 

G 

d 

e 

/ 

FIG.  92. 


FIG.  gza. 


FIG.  93. 


FlG. 


1 76 


APPLIED   MECHANICS. 


§138.  Hammer-Beam  Truss  (Fig.  94). — This  form  of 
truss  is  frequently  used  in  constructions  where  architectural 
effect  is  the  principal  consideration  rather  than  strength.  It 
is  not  an  advantageous  form  from  the  point  of  view  of  strength, 


FIG.  94(5. 


FIG.  94<r. 


for  the  absence  of  a  tie-rod  joining  the  two  lower  joints  causes 
a  tendency  to  spread  out  at  the  base,  which  tendency  is  usually 
counteracted  by  the  horizontal   thrust  furnished  by  the  but 
tresses  against  which  it  is  supported. 


HAMMER-BEAM   TRUSS.  177 

When  such  a  thrust  is  furnished  (or  were  there  a  tie-rod), 
and  the  load  is  symmetrical  and  vertical,  the  bars  are  not  all 
needed,  and  some  of  them  are  left  without  any  stress.  In 
the  case  in  hand,  it  will  be  found  that  UV>  VM,  MQ,  and  QR 
are  not  needed.  We  must  also  observe  that  the  effect  of  the 
curved  members  MY,  MV,  MQ,  and  MN  on  the  other  parts  of 
the  truss  is  just  the  same  as  though  they  were  straight,  as 
shown  in  the  dotted  lines.  The  curved  piece,  of  course,  has  to 
be  subjected  to  a  bending-stress  in  order  to  resist  the  stress 
acting  upon  it.  If,  as  is  generally  the  case,  the  abutments  are 
capable  of  furnishing  all  the  horizontal  thrust  needed,  it  will 
first  be  necessary  to  ascertain  how  much  they  will  be  called 
upon  to  furnish.  To  do  this,  observe  that  we  have  really  a  truss 
similar  to  that  shown  in  Fig.  92,  supported  on  two  inclined 
framed  struts,  of  which  the  lines  of  resistance  are  the  dotted 
lines  (Fig.  94)  i  4  and  7  8,  and  that,  under  a  symmetrical  load, 
this  polygonal  frame  will  be  in  equilibrium,  and,  moreover,  the 
curved  pieces  MV  and  MQ  will  be  without  stress,  these  only 
being  of  use  to  resist  unsymmetrical  loads,  as  the  snow  or 
wind. 

Let  the  whole  load,  concentrated  by  means  of  the  purlins 
at  the  joints  of  the  rafters,  be  W.  Then  will  the  truss  467  have 

to  bear  \  W,  and  this  will  give  —  to  be  supported  at  each  of 

4 
the  points  4  and  7.     Moreover,  on  the  space  2  4  is  distributed 

W 

•— ,  which  has,  as  far  as  overturning  the  strut  is  concerned,  the 

4 

same  effect  as  —  at  2,  and  —  at  4.     Hence  the  load  to  be  sup- 
8  8 

ported  at  4  by  the  inclined  strut  is  a  vertical  load  equal  to 
(\  +  i)  W  —  |  W.  We  may  then  find  the  force  that  must  be 
furnished  by  the  abutment,  or  by  the  tie-rod,  in  either  of  the 
two  following  ways  :  — 


APPLIED  MECHANICS. 


i°.  By  constructing  the  triangle  ySe  (Fig.  94^),  with  yS  = 
|  W,  ye  ||  14,  and  eS  parallel  to  the  horizontal  thrust  of  the  abut- 
ment ;  then  will  ySe  be  the  triangle  of  forces  at  I,  and  eS  will  be 
the  thrust  at  i. 

2°.  Multiply  |  W  by  the  perpendicular  distance  from  4  to 
i  2,  and  divide  by  the  height  of  4  above  I  8  for  the  thrust  of  the 
abutment  ;  in  other  words,  take  moments  about  the  point  i. 

Now,  to  construct  the  diagram  of  stresses,  let,  in  Fig.  94^, 
the  loads  be 

ab,  be,  cd,  de,  ef,  fg,  gh,  kk,  and  klt 
and  let 

lz  =  za  =  \W 

be  the  vertical  component  of  the  supporting  force  ;  let  zm  be 
the  thrust  of  the  abutment  :  then  will  lm  and  ma  be  the  real 
supporting  forces  ;  and  we  shall  have,  for  polygon  of  external 
forces, 

abcdefghklma. 

Then,  proceeding  to  the  joint  i,  we  obtain,  for  polygon  of  forces, 

maym  ; 

and,  proceeding  from  joint  to  joint,  we  obtain  the  stresses  in  all 
the  members  of  the  truss,  as  shown  in  Fig.  94^. 

It  will  be  noticed  that  UV  and  RQ  are  also  free  from 
stress. 

If  we  had  no  horizontal  thrust  from  the  abutment,  and  the 
supporting  forces  were  vertical,  the  members  MV  and  MO 
would  be  called  into  action,  and  MY  and  MN  would  be  inactive. 
To  exhibit  this  case,  I  have  drawn  diagram  94^,  which  shows 
the  stresses  that  would  then  be  developed.  A  Fand  NL  would 
become  merely  part  of  the  supports. 

In  this  latter  case  the  stresses  are  generally  much  greater 
than  in  the  former,  and  a  stress  is  developed  in  UV. 


SC2SSOR-BEAM   TRUSS. 


179 


|  139.  Hammer-Beam  Truss:  Wind  Pressure.  —  Fig.  95 
shows  the  stress  diagram  of  the  hammer-beam  truss  for  wind 
pressure  when  there  is  no  roller  under  either  end,  and  when 
the  wind  blows  from  the  left.  A  similar  diagram  would  give  the 
stresses  when  it  blows  from  the  right. 


FIG.  95- 


FIG.  g$a. 


The  cases  when  there  is  a  roller  are  not  drawn  :  the  student 
may  construct  them  for  himself. 

§140.   Scissor-Beam  Truss.  —  We  have  already  discussed 
two  forms  of  scissor-beam  truss  t 

in  Figs.  90  and  91.  These 
trusses  having  the  right  number 
of  parts,  their  diagrams  present 
no  difficulty.  Another  form  of 
the  scissor-beam  truss  is  shown 
in  Fig.  96,  and  its  diagram  pre- 
sents no  difficulty. 

The  only  peculiarity  to  be  noticed  is,  that,  after  having  con 
structed  the  polygon  of  external  forces, 

abcdefma, 

we  cannot  proceed  to  construct  the  polygon  of  equilibrium  for 
one  of  the  supports,  because  there  are  three  unknown  forces 


i  So 


APPLIED   MECHANICS. 


there.  We  therefore  begin  at  the  apex  CD,  and  construct  the 
triangle  of  forces  cdl  for  this  point ;  then  proceed  to  joint  CB> 
and  construct  the  quadrilateral 

bclkb; 
then  proceed  to  the  left-hand  support,  and  obtain 


and  so  continue. 

§  141.     Scissor-Beam  Truss  without   Horizontal  Tie.  — 

Very  often  the  scissor-beam  truss  is  constructed  without  any 
horizontal  tie,  in  which  case  it  has  the  appearance  of  Fig.  97, 
where  there  is  sometimes  a  pin  at  GKLH  and  sometimes  not. 


FIG.  97. 


FIG.  973. 


FIG.  gjc. 


In  this  case,  if  the  abutments  are  capable  of  furnishing  hori- 
zontal thrust  to  take  the  place  of  the  horizontal  tie  of  Fig.  96, 
we  are  reduced  back  to  that  case.  If  the  abutments  are  not 
capable  of  furnishing  horizontal  thrust,  we  are  then  relying  on 
the  stiffness  of  the  rafters  to  prevent  the  deformation  of  the 
truss;  for,  were  the  points  BC  and  DE  really  joints,  with  pins, 
the  deformation  would  take  place,  as  shown  in  Fig.  97^  or  Fig. 
97$,  according  as  the  two  inclined  ties  were  each  made  in  one 
piece  or  in  two  (i.e.,  according  as  they  are  not  pinned  together 
at  KH,  or  as  they  are  pinned).  This  necessity  of  depending 
on  the  stiffness  of  the  rafters,  and  the  liability  to  deformation 
if  they  had  joints  at  their  middle  points,  become  apparent  as 
soon  as  we  attempt  to  draw  the  diagram.  Such  an  attempt  is 


SCISSOR-BEAM   TRUSS    WITHOUT  HORIZONTAL    TIE.      l8l 

made  in  Fig.  97^,  where  abcdefga  is  the  polygon  of  external 
forces,  gabkg  the  polygon  of  stresses  for  the  left-hand  support, 
kbclk  that  for  joint  BC.  Then,  on  proceeding  to  draw  the  tri- 
angle of  stresses  for  the  vertex,  we  find  that  the  line  joining  d 
and  /  is  not  parallel  to  DL,  and  hence  that  the  truss  is  not 
stable.  We  ought,  however,  in  this  latter  case,  when  the  sup- 
porting forces  are  vertical,  and  when  we  rely  upon  the  stiffness 
of  the  rafters  to  prevent  deformation,  to  be  able  to  determine 
the  direct  stresses  in  the  bars  ;  and  for  this  we  will  employ  an 
analytical  instead  of  a  graphical  method,  as  being  the  most  con- 
venient in  this  case. 

Let  us  assume  that  there  is  no  pin  at  the  intersection  of  the 
two  ties,  and  that  the  two  rafters  are  inclined  at  an  angle  of  45° 
to  the  horizon. 

We  then  have,  if  W  =.  the  entire  load,  and  a  =  angle 
between  BK  and  KG, 

W  W 

ab  =  ef  =  — ,      be  =  cd  =  de  =  — , 
8  4 

tana  =  £,     sin  a  =  -p,     cos  a  =  -p, 

Let  x  be  the  stress  in  each  tie,  and  let  y  =  cl  =  dl  =  thrust 
in  each  upper  half  of  the  rafters. 

Then  we  must  observe  that  the  rafter  has,  in  addition  to  its 
direct  stresses,  a  tendency  to  bend,  due  to  a  normal  load  at  the 
middle,  this  normal  load  being  equal  to  the  sum  of  the  normal 
components  of  be  and  of  x,  when  these  are  resolved  along  and 
normal  to  the  rafter.  Hence 

normal  load  =  x  cos  a  H sin  45°. 

4 

This,  resolved  into  components  acting  at  each  end  of  the  rafter, 
gives  a  normal  downward  force  at  each  end  equal  to 


1  82  APPLIED   MECHANICS. 

Hence,  resolving  all  the  forces  acting  at  the  left-hand  support 
into  components  along  and  at  right  angles  to  the  rafter,  and 
imposing  the  condition  of  equilibrium  that  the  algebraic  sum 
of  their  normal  components  shall  equal  zero,  we  have,  if  we  call 
upward  forces  positive, 

|^sin45°  -  (£#cosa  +  |£Fsin45°)  -  *sma  =  o;     (i) 

but,  since 

.*  cos  a  =  2x  sin  a, 
we  have  from  (i) 

2X  sin  a  =  —  sin  45° 
4 

W 

/.    #sma    =  —  sin  45 
8 


sn  a 
Then,  proceeding  to  the  apex  of  the  roof,  we  have  that  the  load 

«?,£ 

4 

gives,  when  resolved  along  the  two  rafters,  a  stress  in  each 
equal  to 


4 

Hence  the  load  to  be  supported  in  a  direction  normal  to  the 
rafter  at  the  apex  is 

-sin  45°  +  (i*cosa  +  ^sm45°). 
4  8 

Hence,  substituting  for  x  its  value,  we  have 

j,  =  ,/=,#=  ^sin  45°.  (3) 

Then,  proceeding  to  the  left-hand  support,  and  equating  to  zero 
the  algebraic  sum  of  the  components  along  the  rafter,  we  have 

bk  =  (ga  —  <z^)cos45°  •+•  ^cosa 

=  f  ^sin  45°  +  i  ^sin  45°  =  f  ^sin  45°.        (4) 


SCISSOR-BEAM   TRUSS   WITHOUT  HORIZONTAL    TIE.      183 


We  have  thus  determined  in  (2),  (3),  and  (4)  the  values  of  x,  y, 
and  bk  =  eh. 

By  way  of  verification,  proceed  to  the  middle  of  the  left- 
hand  rafter,  and  we  find  the  algebraic  sum  of  the  components 


of  be  and  x  along  the  rafter  to  be 


and  this  is  the  difference  between  bk  and  cl,  as  it  should  be. 

We  have  thus  obtained  the  direct  stresses  ;  and  we  have,  in 
addition,  that  the  rafter  itself  is  also  subjected  to  a  bending- 
moment  from  a  normal  load  at  the  centre,  this  load  being  equal 

to 

W  W 

xcosa  H  --  sin  45    =   —sin  45°. 
4  2 

How  to  take  this  into  account  will  be  explained  under  the 
"  Theory  of  Beams." 

.  §142.  Examples.  —  The  following  figures  of  roof-trusses 
may  be  considered  as  a  set  of  examples,  for  which  the  stress 
diagrams  are  to  be  worked  out. 

Observe,  that,  wherever  there  is  a  joint,  the  truss  is  to  be 
supposed  perfectly  flexible,  i.e.,  free  to  turn  around  a  pin. 


FIG.  98. 


FIG.  99. 


FIG. 


FIG.  i 01. 


FIG.  102. 


FIG.  103. 


FIG.  104. 


FIG.  105. 


FIG.  106. 


FIG. 


FIG.  108 


TS4  APPLIED   MECHANICS. 


CHAPTER   IV. 
BRIDGE-TRUSSES. 

§  143.  Method  of  Sections.  —  It  is  perfectly  possible  to 
determine  the  stresses  in  the  members  of  a  bridge-truss 
graphically,  or  by  any  methods  that  are  used  for  roof-trusses. 

In  this  work  an  analytical  method  will  be  used  ;  i.e.,  a  method 
of  sections.  This  method  involves  the  use  of  the  analytical  con- 
ditions of  equilibrium  for  forces  in  a  plane  explained  in  §  63. 
These  are  as  follows  ;  viz.,  — 

If  a  set  of  forces  in  a  plane,  which  are  in  equilibrium,  be 
resolved  into  components  in  two  directions  at  right  angles  to 
each  other,  then  — 

i°.  The  algebraic  sum  of  the  components  in  one  of  these 
directions  must  be  zero. 

2°.  The  algebraic  sum  of  the  components  in  the  other  of 
these  directions  must  be  zero. 

3°.  The  algebraic  sum  of  the  moments  of  the  forces  about 
any  axis  perpendicular  to  the  plane  of  the  forces  must  be  zero. 

Assume,  now,  a  bridge-truss  (Figs.  109,  no,  in,  112,  pages 
176  and  177)  loaded  at  a  part  or  all  of  the  joints.  Conceive  a 
vertical  section  ab  cutting  the  horizontal  members  6-8  and  7-9 
and  the  diagonal  7-8,  and  dividing  the  truss  into  two  parts. 
Then  the  forces  acting  on  either  part  must  be  in  equilibrium, 
in  other  words,  fhe  external  forces,  loads,  and  supporting  forces, 
acting  on  one  part,  must  be  balanced  by  the  stresses  in  the 
members  cut  by  the  section ;  i.e.,  by  the  forces  exerted  by  the 
other  part  of  the  truss  on  the  part  under  consideration.  Hence 
we  must  have  the  three  following  conditions ;  viz.,  — 


SHEARING-FORCE   AND  BEND  ING-MOMENT.  185 

i°.  The  algebraic  sum  of  the  vertical  components  of  the 
above-mentioned  forces  must  be  zero. 

2°.  The  algebraic  sum  of  the  horizontal  components  of  these 
forces  must  be  zero. 

3°.  The  algebraic  sum  of  the  moments  of  these  forces  about 
any  axis  perpendicular  to  the  plane  of  the  truss  must  be  zero. 

§  144.  Shearing-Force  and  Bending-Moment.  —  Assum- 
ing all  the  loads  and  supporting  forces  to  be  vertical,  we  shall 
have  the  following  as  definitions. 

The  Shearing-Force  at  any  section  is  the  force  with  which 
the  part  of  the  girder  on  one  side  of  the  section  tends  to  slide 
by  the  part  on  the  other  side. 

In  a  girder  free  at  one  end,  it  is  equal  to  the  sum  of  the 
loads  between  the  section  and  the  free  end. 

In  a  girder  supported  at  both  ends,  it  is  equal  in  magnitude 
to  the  difference  between  the  supporting  force  at  either  end, 
and  the  sum  of  the  loads  between  the  section  and  that  support- 
ing force. 

The  Bending-Moment  at  any  section  is  the  resultant  moment 
of  the  external  forces  acting  on  the  part  of  the  girder  to  one  side 
of  the  section,  tending  to  rotate  that  part  of  the  girder  around 
a  horizontal  axis  lying  in  the  plane  of  the  section. 

In  a  girder  free  at  one  end,  it  is  equal  to  the  sum  of  the 
moments  of  the  loads  between  the  section  and  the  free  end, 
about  a  horizontal  axis  in  the  section. 

In  a  girder  supported  at  both  ends,  it  is  the  difference  be- 
tween the  moment  of  either  supporting  force,  and  the  sum  of 
the  moments  of  the  loads  between  the  section  and  that  sup- 
port ;  all  the  moments  being  taken  about  a  horizontal  axis  in 
the  section. 

§  145.  Use  of  Shearing-Force  and  Bending-Moment.  — 
The  three  conditions  stated  in  §  143  may  be  expressed  as  fol- 
lows :  — 

i°.  The  algebraic  sum  of  the  horizontal  components  of  the 
stresses  in  the  members  cut  by  the  section  must  be  zero. 


1 36 


APPLIED   MECHANICS. 


2°.  The  algebraic  sum  of  the  vertical  components  of  the 
stresses  in  the  members  cut  by  the  section  must  balance  the 
shearing-force. 

3°.  The  algebraic  sum  of  the  moments  of  the  stresses  in 
the  members  cut  by  the  section,  about  any  axis  perpendicular  to 
the  plane  of  the  truss,  and  lying  in  the  plane  of  the  section, 
must  balance  the  bending-moment  at  the  section. 

As  the  conditions  of  equilibrium  are  three  in  number,  they 
will  enable  us  to  determine  the  stresses  in  the  members,  pro- 
vided the  section  does  not  cut  more  than  three ;  and  this 
determination  will  require  the  solution  of  three  simultaneous 
equations  of  the  first  degree  with  three  unknown  quantities 
(the  stresses  in  the  three  members). 

By  a  little  care,  however,  in  choosing  the  section,  we  can 
very  much  simplify  the  operations,  and  reduce  our  work  to  the 
solution  of  one  equation  with  only  one  unknown  quantity  ;  the 
proper  choice  of  the  section  taking  the  place  of  the  elimination. 

§  146.  Examples  of  Bridge-Trusses.  —  Figs.  109-112  rep- 
resent two  common  kinds  of  bridge-trusses  :  in  the  first  two 

the  braces  are  all 
diagonal,  in  the 
last  two  they  are 
partly  vertical  and 
FlG-  I09'  partly  diagonal. 

The  first  two  are  called  Warren  girders,  or  half-lattice  girders ; 
since  there  is  only  one  system  of  bracing, 
as  in  the  figures.  When,  on  the  other 
hand,  there  are  more  than  one  system,  so 
that  the  diagonals  cross  each  other,  they 
are  called  lattice  girders. 

§  147.  General  Outline  of  the  Steps 
to  be  taken  in  determining  the  Stresses 
in  a  Bridge-Truss  under  a  Fixed  Load. 

i°.  If  the  truss  is  supported  at  both  ends,  find  the  sup- 
porting forces. 


5      7 


IV\/VN 


2466 


11     13     45     17    19    21 


25    27 


VVVXAAA/VW 


10     12     14     16     18     20     22     24 


1357 

a  <) 

xll      13 

vw 

2 

K/V 

2466 

8  z 

10     13 

FIG.  no. 


DETERMINING    THE   STRESSES  IN  A    BRIDGE-TRUSS.   l8/ 


2°.  Assume,  in  all  cases,  a  section,  in  such  a  manner  as  not 
to  cut  more  than  three  members  if  possible,  or,  rather,  three 


oi  those  mat 
are   brought 
into     action 
by  the  loads 
on  the  truss  ; 
and    it    will 

1        3       5       7a 

0 

11      13      15      17      19     21     23      25      27      28 

I 

\l\ 

\is 

\ 

^XK 

ymAAAA/ 

I 

24       66 

8 

10      12      1 
FIG. 

4     16     18     20     22     24      2G 
in. 

2      4       6 

8 

10    12    1 

AAA, 

4 

/I/I/ 

1357 

9 

11    13 

FIG.  112. 


save  labor  if  we  assume  the  section  so  as  to  cut  two  of  the 

three  very  near  their  point  of  inter- 
section. 

3°.  Find  the  shearing-force  at  the 
section. 

4°.  Find   the   bending-moment   at 
the  section. 

5°.  Impose  the  analytical  conditions  of  equilibrium  on  all 
the  forces  acting  on  the  part  of  the  girder  to  one  side  of  the 
section,  —  the  part  between  the  section  and  the  free  end  when 
the  girder  is  free  at  one  end,  or  either  part  when  it  is  supported 
at  both  ends. 

In  the  cases  shown  in  Figs.  109  and  no,  we  may  describe 
the  process  as  follows  ;  viz.,  — 

(a)  Find  the  stress  in  the  diagonal  from  the  fact,  that  (since 
the  stress  in  the  diagonal  is  the  only  one  that  has  a  vertical 
component  at  the  section)  the  vertical  component  of  the  stress 
in  the  diagonal  must  balance  the  shearing-force. 

(b]  Take  moments  about  the  point  of  intersection  of  the 
diagonal  and  horizontal  chord  near  which  the  section  is  taken  ; 
then  the  stresses  in  those  members  will  have  no  moment,  so 
that   the  moment    of   the   stress  in  the  other  horizontal  must 
balance  the  bending-moment  at  the  section.     Hence  the  stress 
in  the  horizontal  will  be  found  by  dividing  the  bending-moment 
at  the  section  by  the  height  of  the  girder. 

The  above  will  be  best  illustrated  by  some  examples. 


1 88  APPLIED   MECHANICS. 

EXAMPLE  I.  —  Given  the  semi-girder  shown  in  Fig.  no, 
loaded  at  joint  13  with  4000  pounds,  and  at  each  of  the  joints 
*»  3>  5>  7)  9>  and  ii  with  8000  pounds.  Suppose  the  length  of 
each  chord  and  each  diagonal  to  be  5  feet.  Required  the  stress 
in  each  member. 

Solution.  —  For  the  purpose  of  explaining  the  method  of 
procedure,  we  will  suppose  that  we  desire  to  find  first  the 
stresses  in  8-ip  and  9-10. 

Assume  a  vertical  section  very  near  the  joint  9,  but  to  the 
right  of  it,  so  that  it  shall  cut  both  8-10  and  9-10. 

If,  now,  the  truss  were  actually  separated  into  two  parts  at 
this  section,  the  right-hand  part  would,  in  consequence  of  the 
loads  acting  on  it,  separate  from  the  other  part.  This  tendency 
to  separate  is  counteracted  by  the  following  three  forces  :  — 

i°.  The  pull  exerted  by  the  part  g-x  of  the  bar  9-11  on  the 
part  x-\\  of  the  same  bar. 

2°.  The  thrust  exerted  by  the  part  8-z  of  the  bar  8-10  on 
the  part  z-io  of  tne  same  bar. 

3°.  The  pull  exerted  by  the  part  q-y  of  the  bar  9-10  on  the 
part  y-io  of  the  same  bar. 

The  shearing-force  at  this  section  is 

8000  -\-  4000  =  12000  Ibs., 

and  this  is  equal  to  the  vertical  component  of  the  stress  in-  the 
diagonal.     Hence 

Stress  in  9-10  =  I.2OO°  =  12000(1.1547)  =  13856  Ibs. 
sin  60 

This  stress  is  a  pull,  as  may  be  seen  from  the  fact,  that,  in 
order  to  prevent  the  part  of  the  girder  to  the  right  of  the 
section  from  sliding  downwards  under  the  action  of  the  load, 
the  part  g-y  of  the  diagonal  9-10  must  pull  the  part  y-\o  of 
the  same  diagonal. 

Next  take  moments  about  9 :  and,  since  the  moment  of  the 
stresses  in  9-1 1  and  9-10  about  9  is  zero,  we  must  have  that  the 
moment  of  the  stress  in  8-10;  i.e.,  the  product  of  this  stress 
by  the  height  of  the  girder,  must  equal  the  bending-moment. 


DETERMINING    THE  STRESSES  IN  A   BRIDGE-TRUSS.    I 


The  bending-moment  about  9  is 

8000  x  5  4-  4000  x  10  =  80000  foot-lbs. 


Hence 


Stress  in  8-10  = 


80000 
4-33 


=  80000(0.23094)  =  18475 


Proceed  in  a  similar  way  for  all  the  other  members.  The 
work  may  be  arranged  as  in  the  following  table ;  the  diagonal 
stresses  being  deduced  from  the  shearing-forces  by  multiplying 
by  1.1547,  and  the  chord  stresses  from  the  bending-moments 
by  multiplying  by  0.23094. 


Section  just  to 
the  right  of 

Shearing- 
Force 

in  Ibs. 

Stresses  in  Diagonals  cut 
by  Section;,  en  Ibs. 

Bending- 
Moment,  in 
foot-lbs. 

Stresses  in  Chords  opposite  the 
respective  Joints. 

Tension. 

Compression. 

Tension. 

Compression. 

I 

44OOO 

50806 

720000 

166277 

2 

44OOO 

50806 

6lOOOO 

140873 

3 

36000 

4^9 

5OOOOO 

II5470 

4 

36000 

41569 

4IOOOO 

94685 

5 

28OOO 

32331 

32OOOO 

73901 

6 

28000 

32331    }    250000 

57735 

7 

2OOOO 

23094 

\ 

180000 

41569 

8 

2OOOO 

23094 

I3OOOO 

30022  | 

9 

I20OO 

13856 

8OOOO 

18475 

10 

I20OO 

13856 

500OO 

H547 

ii 

4OOO 

4619 

2OOOO 

4618 

12 

4OOO 

4619 

IOOOO 

2309 

EXAMPLE  II.  —  Given  the  truss  (Fig.  109)  loaded  at  each  of 
the  lower  joints  with  10000  Ibs.  :  find  the  stresses  in  the  members. 
The  length  of  chord  is  equal  to  the  length  of  diagonal  =  10  ft. 

In  future,  tensions  will  be  written  with  the  minus,  and  com- 
pressions with  the  plus  sign. 

Solution.  —  Total  load  —  14(10000)  =  140000  Ibs. 
Each  supporting  force        =     70000    " 

The  entire  work  is  shown  in  the  following  tables  :  — - 


1 90 


APPLIED   MECHANICS. 


O\      M 
II        II 


O      00 


N       M       M 
II        II        II 


CO      ^T      CO      Tj-      CO 


O      LO 
CO      M 


w->     O       in     O      *0      O       LT>O 
_n       «n^M«M 

LT.Q          ^O          LOO          L^O 

oooooooo 


I    I    I    I    I    I    I    I    I    I    I    I 

O>j-.     OLnOmO^OLOOLr, 

r-l  (_i  (VJ  M  COCO^VfU^Ln^V^ 

XXXXXXXXXXXX 

oooooooooooo 
oooooooooooo 


8 


1          1          1          1 


»-   O   CN  oo   i^  O   I 


O   -.   N   CO 


DETERMINING    THE  STRESSES  IN  A    BRIDGE-TRUSS. 


X 

Numbers  of  Diagonals. 

Stresses  in  Diagonals,  in 

Ibs. 

I-    2 

28-29 

—  70000  x  1.1547  = 

-80829 

2-  3 

27-28 

+60000  x  1.1547  = 

+69282 

3-  4 

26-27 

—60000  x  1.1547  = 

—69282 

4-  5 

25-26 

+50000  x  1.1547  = 

+  57735 

5-  6 

24-25 

—50000  x  1.1547  = 

-57735 

6-  7 

23-24 

+40000  x  1.1547  = 

+46188 

7-  8 

22-23 

—40000  x  1.1547  = 

—  46188 

8-  9 

21-22 

+  30000  X    1.1547  = 

+  34641 

9-10 

2O-2I 

—  30000   X    1.1547  = 

—  34641 

IO-II 

I9-2O 

+  20000   X    1.1547  = 

+  23094 

11-12 

18-19 

—  20000  X   1.1547  = 

-23094 

12-13 

I7-I8 

+  10000  X   1.1547  = 

+  H547 

I3-M 

16-17 

—  10000  x  1.1547  = 

-H547 

14-15 

I5-I6 

+  0 

0 

LOWER    CHORDS. 


Numbers  of  Chords. 

Stresses 

in  Chords,  in 

Ibs. 

2-  4 

26-28 

—      650000    X    O. 

"547  = 

'     75056 

4-  6 

24-26 

—    I2OOOOO    X    0. 

"547  = 

-138564 

6-  8 

22-24 

—  1650000  X  o. 

"547  = 

—  190526 

8-10 

20-22 

—   2000000    X    0. 

H547  = 

-230940 

10-12 

I  8-20 

—    225OOOO    X    O. 

"547  = 

—  259808 

12-14 

16-18 

—  2450000  X  o. 

"547  = 

—  277128 

I4-l6 

—  2450000   X   O. 

"547  = 

—  282902 

1 92 


APPLIED   MECHANICS. 


UPPER    CHORDS. 


Numbers  of  Chords. 

Stresses  in  Chords,  in  Ibs. 

!-  3 

27-29 

350000 

X   0.11547  = 

+  40415 

3-  5 

25-27 

95OOOO 

X   0.11547  = 

+  109697 

5-  7 

23-25 

I45OOOO 

X   0.11547  = 

+  167432 

7-  9 

21-23 

1850000 

X   0.11547  = 

+  213620 

9-1  1 

19-21 

2I5OOOO 

X   0.11547  = 

+  248261 

11-13 

17-19 

235OOOO 

X   0.11547  = 

+  267355 

13-15 

I5-I7 

245OOOO 

X   0.11547  = 

+  282902 

EXAMPLE  III.  —  Given  the  same  truss  as  in  Example  II., 
loaded  ^at  2,  4,  6,  8,  10,  and  12  with  10000  Ibs.  at  each  point, 
the  remaining  lower  joints  being  loaded  with  5000x3  Ibs.  at  each 
joint :  find  the  stresses  in  the  members. 

EXAMPLE  IV.  —  Given  a  semi-girder,  free  at  one  end  (Fig. 
112),  loaded  at  2,  4,  and  6  with  10000  Ibs.,  and  at  8,  10,  and  12 
with  5000  Ibs. :  find  the  stresses  in  the  members. 

TRAVELLING-LOAD. 

§148.  Half-Lattice  Girder:  Travelling-Load.  —  When  a 
girder  is  used  for  a  bridge,  it  is  not  subjected  all  the  time  to 
the  same  set  of  loads. 

The  load  in  this  case  consists  of  two  parts, — one,  the  dead 
load,  including  the  bridge  weight,  together  with  any  permanent 
load  that  may  rest  upon  the  bridge  ;  and  the  other,  the  moving 
or  variable  load,  also  called  the  travelling-load,  such  as  the 
weight  of  the  whole  or  part  of  a  railroad  train  if  it  is  a  railroad 
bridge,  or  the  weight  of  the  passing  teams,  etc.,  if  it  is  a  common- 
road  bridge.  Hence  it  is  necessary  that  we  should  be  able  to 
determine  the  amount  and  distribution  of  the  loads  upon  the 
bridge  which  will  produce  the  greatest  tension  or  the  greatest 


GREATEST  DIAGONAL   STRESSES  IN  GIRDER.  193 

compression  in  every  member,  and  the  consequent  stress  pro- 
duced. 

§149.  Greatest  Stresses  in  Semi-Girder.  —  Wherever  the 
section  be  assumed  in  a  semi-girder,  it  is  evident  that  any  load 
placed  on  the  truss  at  any  point  between  the  section  and  the 
free  end  increases  both  the  shearing-force  and  the  bending- 
moment  at  that  section,  and  that  any  load  placed  between  the 
section  and  the  fixed  end  has  no  effect  whatever  on  either 
the  shearing-force  or  the  bending-moment  at  that  section. 

Hence  every  member  of  a  semi-girder  will  have  a  greater 
stress  upon  it  when  the  entire  load  is  on,  than  with  any  partial 
load. 

§  150.  Greatest  Chord  Stresses  in  Girder  supported  at 
Both  Ends.  —  Every  load  which  is  placed  upon  the  truss,  no 
matter  where  it  is  placed,  will  produce  at  any  section  whatever  a 
bending-moment  tending  to  turn  the  two  parts  of  the  truss  on 
the  two  sides  of  the  section  upwards  from  the  supports  ;  i.e.,  so 
as  to  render  the  truss  concave  upwards. 

Hence  every  load  that  is  placed  upon  the  truss  causes  com- 
pression in  every  horizontal  upper  chord,  and  tension  in  every 
horizontal  lower  chord.  Hence,  in  order  to  obtain  the  greatest 
chord  stresses,  we  assume  the  whole  of  the  moving  load  to  be 
upon  the  bridge. 

§  151.  Greatest  Diagonal  Stresses  in  Girder  supported 
at  Both  Ends.  —  To  determine  the  distribution  of  the  load 
that  will  produce  the  greatest  stress  of  a  certain  kind  (tension 
or  compression)  in  any  given  diagonal,  let  us  suppose  the  diag- 
onal in  question  to  be  7-8  (Fig.  109),  through  which  we  take 
our  section  ab.  Now,  it  is  evident  that  any  load  placed  on  the 
truss  between  ab  and  the  left-hand  (nearer)  support  will  cause  a 
shearing-force  at  that  section  which  will  tend  to  slide  the  part 
of  the  girder  to  the  left  of  the  section  downwards  with  refer- 
ence to  the  other  part,  and  hence  will  cause  a  compressive 
stress  in  7-8  ;  while  any  load  between  the  section  and  the  right- 


194  APPLIED   MECHANICS. 

hand  (farther)  support  will  cause  a  shearing-force  of  the  oppo- 
site kind,  and  hence  a  tension  in  the  bar  7-8. 

Now,  the  bridge  weight  itself  brings  an  equal  load  upon  each 
joint ;  hence,  when  the  bridge  weight  is  the  only  load  upon  the 
truss,  the  bar  7-8  is  in  tension. 

Hence,  any  load  placed  upon  the  truss  between  the  section 
and  the  farther  support  tends  to  increase  the  shearing-force  at 
that  section  due  to  the  dead  load  (provided  this  is  equally  dis- 
tributed among  the  joints) ;  whereas  any  load  placed  between 
the  section  and  the  nearer  support  tends  to  decrease  the  shear- 
ing-force at  the  section  due  to  the  dead  load,  or  to  produce  a 
shearing-force  of  the  opposite  kind  to  that  produced  by  the  dead 
load  at  that  section. 

Hence,  if  we  assume  the  dead  load  to  be  equally  distributed 
among  the  joints,  we  shall  have  the  two  following  propositions 
true :  — 

(a)  In  order  to  determine  the  greatest  stress  in  any  diagonal 
which  is  of  the  same  kind  as  that  produced  by  the  dead  load, 
we  must  assume  the  moving  load  to  cover  all  the  panel  points 
between  the  section  and  the  farther  abutment,  and  no  other 
panel  points. 

(b)  In  order  to  determine  the  greatest  stress  in  any  diagonal 
of  the  opposite  kind  to  that  produced  by  the  dead  load,  we  must 
assume  the  moving  load  to  cover  all  the  panel  points  between 
the  section  and  the  nearer  abutment,  and  no  others. 

This  will  be  made  clear  by  an  example. 

EXAMPLE  I.  —  Given  the  truss  shown  in  Fig.  113.     Length 

of  chord  =  length  of  diagonal  = 

~\  A  A  A  A  A  A  A  %=#    IO  feet'     Dead  load  =  8oo°  lbs' 

3   applied  at  each  upper  panel  point. 

Moving  load  =  30000  Ibs.  applied 
at  each  upper  panel  point.     Find 


2      4       6       8      10     12     14     16 
FIG.  113. 


the  greatest  stresses  in  the  members. 


EXAMPLE   OF  BRIDGE-TRUSS. 


195 


(U  •+•  76788  (3)  *  208423  (5)  +  296181  (7)-+  340059  (9) 


Solution,  (a)  Chord  Stresses.  —  Assume  the  whole  load  to 
be  upon  the  bridge : 
this  will  give  38000 
Ibs.  at  each  upper 
panel  point ;  i.e.,  omit- 
ting I  and  17,  where 
the  load  acts  directly 
on  the  support,  and 
not  on  the  truss. 
Hence,  considering  the  bridge  so  loaded,  we  shall  have  the  fol- 
lowing results  for  the  chord  stresses  :  — 


(2)    -153575  (4)    -263272(6)  -329090  (8)~ 351029 


FIG.  114. 


Each  supporting  force  =  38000! -)  =  133000. 

\2/ 


Section  at 

Bending-Moment,  in  foot-lbs* 

2 

16 

133000  x  5 

=  665000 

3 

J5 

133000  X  10 

=  1330000 

4 

M 

133000  X  15 

-  38000  x  5 

=  1805000 

5 

!3 

133000  X  20 

—  38000  X  IO 

=  2280000 

6 

12 

133000  X  25 

-  38ooo(  5  +  15) 

=  2565000 

7 

II 

133000  X  30 

—  38000(10  +  20) 

=  2850000 

8 

IO 

133000  x  35 

-  38ooo(  5  +  15  +  25) 

=  2945000 

9 

133000  X  40 

—  38000(10  4-  20  -f-  30) 

=  3040000 

Numbers  of  Chords. 

Stresses  in  Upper 

Chords. 

!~3 

I5-I7 

665000   X   0.11547 

=    +    76788 

3-5 

I3-I5 

1805000   X  0.11547 

=    +208423 

5-7 

11-13 

2565000   X   0.11547 

=    +296181 

7-9 

9-1  1 

2945000   X   0.11547 

=    +340059 

APPLIED  MECHANICS. 


Numbers  of  Chords. 

Stresses 

in  Lower  Chords. 

2-  4 

14-16 

—  1330000   X 

O.II547    = 

-153575 

4-  6 

12-14 

—  2280000  X 

O.H547    = 

—  263272 

6-  8 

10-12 

—  2850000   X 

0.11547    = 

—  329090 

8-10 

—  3040000   X 

O.II547    = 

-351029 

Next,  as  to  the  diagonals,  take,  for  instance,  the  diagonal 
7-8.  When  the  dead  load  alone  is  on  the  bridge,  the  diagonal 
7-8  is  in  tension.  From  the  preceding,  we  see  that  the  greatest 
tension  is  produced  in  this  bar  when  the  moving  load  is  on  the 
points  9, ii,  13,  and  15,  and  the  dead  load  only  on  the  points  3, 
5,  7.  Now,  a  load  of  38000  Ibs.  at  13,  for  instance,  causes  a 

shearing-force  of  —(38000)  -=  9500  Ibs.  at  any  section  to  the 

left  of  1 3  ;  and  this  shearing-force  tends  to  cause  the  part  to 
the  left  of  the  section  to  slide  upwards,  and  that  to  the  right 
downwards. 

On  the  other  hand,  with  the  same  load  at  the  same  place, 

there  is  produced  a  shearing-force  of  —(38000)  =  28500    Ibs. 

ID 

at  any  section  to  the  right  of  13  ;  and  this  shearing-force  tends 
to  cause  the  part  to  the  left  to  slide  downwards,  and  that  to  the 
right  upwards.  Paying  attention  to  this  fact,  we  shall  have, 
when  the  loads  are  distributed  as  above  described,  a  shearing- 
force  at  the  bar  7-8  causing  tension  in  this  bar ;  the  magnitude 
of  this  shearing-force  being 


16 


8ooo 
~i6~ 


(2 


41500. 


Hence,  we  may  arrange  the  work  as  follows  :  — 


GREATEST  DIAGONAL   STRESSES  IN  GIRDER. 


197 


Greatest 

Stresses  in 

Numbers  of 
Diagonals. 

Greatest  Shearing-Forces  producing  Stresses  of  Same  Kind  as 
Dead  Load. 

Diagonals  of 
Same  Kind 
as  those  due 

to  Dead 

Load. 

1-2 

17-16 

^-^(2  +  4+6+8+10+12+14)                            =  133000 

-153575 

2-3 

16-15 

^^(2+4+6+8+10+12+14)                            =  133000 

+  153575 

3-4 

15-14 

^(2+4+6+8+10+12)  -^(2)              =    98750 

-114027 

4-5 

14-13 

^(2+4+6+8+10+12)  -^(2)              =    98750 

+  114027 

5-6 

13-12 

^(2+4+6+8+10)          -^(2+4)        =    68250 

-  78808 

6-7 

1  2-1  1 

3-^P(2+4+6+8+IO)          -^(2+4)        =    68250 

+   78808 

7-8 

II-IO 

^(2+4+6+8)                 -  ^(2+4+6)  =    41500 

-  47920 

8-9 

IO—  9 

^(2+4+6+8)                 -  ^(2+4+6)  -    41500 

+   47920 

Greatest 

Stresses  in 

Numbers  of 

Greatest  Shearing-Forces  producing  Stresses  of  Kind  Opposite 

Diagonals  of 

Diagonals. 

from  Dead  Load. 

Kind  Oppo- 

site from 

Dead  Load. 

8-9 

io-  9 

^(2+4+6)  -  ^(2+4+6+8)                   =  18500 

—21362 

7-8 

II-IO 

^(2+4+6)  -  ^(2+4+6+8)                   =  18500 

+  21362 

The  diagonals  7-8,  8-9,  9-10,  and  10-11  are  the  only  ones 
that,  under  any  circumstances,  can  have  a  stress  of  the  kind 
opposite  to  that  to  which  they  are  subjected  under  the  dead 
load  alone. 


APPLIED  MECHANICS. 


Fig.  114  exhibits  the  manner  of  writing  the  stresses  on  the 
diagram. 

§152.  General  Application  of  this  Method.  —  It  is  plain 
that  the  method  used  above  will  apply  to  any  single  system  of 
bridge-truss  with  horizontal  chords  and  diagonal  bracing,  what- 
ever be  the  inclination  of  the  braces. 

When  seeking  the  stress  in  a  diagonal,  the  section  must  be 
so  taken  as  to  cut  that  diagonal  ;  and,  as  far  as  this  stress  alone 
is  concerned,  it  may  be  equally  well  taken  at  any  point,  as  well 
as  near  a  joint,  provided  only  it  cuts  that  diagonal  which  is  in 
action  under  the  load  that  produces  the  greatest  stress  in  this 
one,  and  no  other. 

On  the  other  hand,  when  we  seek  the  stress  in  a  horizontal 
chord,  the  section  might  very  properly  be  taken  through  the 
joint  opposite  that  chord. 

Taking  it  very  near  the  joint,  only  serves  to  make  one  sec- 
tion answer  both  purposes  simultaneously. 

§  153.  Bridge-Trusses  with  Vertical  and  Diagonal  Bra- 
cing. —  When,  as  in  Figs,  in  and  112,  there  are  both  vertical 
and  diagonal  braces,  and  also  horizontal  chords,  we  may  deter- 
mine the  stresses  in  the  diagonals  and  in  the  chords  just  as 
before  ;  only  we  must  take  the  section  just  to  one  side  of  a  joint, 
and  never  through  the  joint. 

As  to  the  verticals,  in  order  to  determine  the  stress  in  any 
vertical,  we  must  impose  the  conditions  of  equilibrium  between 
the  vertical  components  of  the  forces  acting  at  one  end  of  that 
vertical:  thus,  if  the  loads  are  at  the  upper  joints  in  Fig.  in, 
then  the  stress  in  vertical  3-2  must  be  equal  and  opposite  to 
the  vertical  component  of  the  stress  in  diagonal  1-2,  as  these 
stresses  are  the  only  vertical  forces  acting  at  joint  2. 

Vertical  5-4  has  for  its  stress  the  vertical  component  of  the 
stress  in  3-4,  etc.  Thus 

Stress  in  3-2  =  shearing-force  in  panel  1-3, 
Stress  in  5-4  =  shearing-  force  in  panel  3-5,  etc. 


TRUSSES   WITH   VERTICAL   AND  DIAGONAL   BRACING.    1 99 

On  the  other  hand,  if  the  loads  be  applied  at  the  lower 
joints,  then 

Stress  in  3-2  =  shearing-force  in  panel  3-5, 
Stress  in  5-4  =  shearing-force  in  panel  5-7,  etc. 

EXAMPLE.  —  Given  the  truss  shown  in  Fig.  in.  Given 
panel  length  =  height  of  truss  =  10  feet,  dead  load  per  panel 
point  =  12000  Ibs.,  moving  load  per  panel  point  =  23000  Ibs.  ; 
load  applied  at  upper  joints. 

Solution,  (a)  Chord  Stresses.  —  Assume  the  entire  load  on 
the  bridge,  i.e.,  35000  Ibs.  per  panel  point.  Hence 

Total  load  on  truss  =  13  (35000)  =  455000  Ibs., 
Each  supporting  force  =  227500  Ibs. 


Joint  near 

which 
Section  is 

Bending-Moment  at  the  Section  very  near  the  Joint,  on 

Either  Side  of  the  Joint. 

taken. 

I        28 

0 

3    ,27 

227500  x  10 

=  2275000 

5     25 

227500  X   20  —  350OO  X   IO 

=  42OOOOO 

7     23 

227500  X  30  —  35000(10  +  20) 

=  5775000 

9     21 

227500  x  40  —  35000  (10  +  20  +  30) 

=  7000000 

n      19 

227500  X   50  —  35000  (10  +  20  +  30  +  40) 

=  7875000 

13      17 

227500  X  60  —  35000(lO  +  20  +  30  +  40  + 

50)              =  8400000 

15 

227500  X   70  —  35000  (10  +  20  +  30  +  40  + 

50  +  60)  =  8575000 

To  find  any  chord  stress,  divide  the  bending-moment  at  a 
section  cutting  the  chord,  and  passing  close  to  the  opposite 
joint,  by  the  height  of  the  girder,  which  in  this  case  is  10. 
Hence  we  have  for  the  chord  stresses  (denoting,  as  before,  com- 
pression by  +,  and  tension  by  — ) :  — 


200 


APPLIED   MECHANICS. 


Stresses  in  Upper  Chords. 

Stresses  in  Lower  Chords. 

J-  3 

27-28 

4-227500 

2-  4 

24-26 

—  227500 

3-  5 

25-27 

+420000 

4-  6 

22-24 

—  420000 

5-  7 

23-25 

+5775°° 

6-  8 

20-22 

-5775°o 

7-  9 

21-23 

+  700000 

8-10 

1  8-20 

—  700000 

9-1  1 

19-21 

+  787500 

IO-I2 

1  6-1  8 

-787500 

11-13 

17-19 

+  840000 

12-14 

14-16 

—  840000 

*3-i5 

iJT:?.? 

+  857500 

Diagonals.  —  It  is  evident,  that,  for  the  diagonals,  the  same 
rule  holds  as  in  the  case  of  the  Warren  girder  :  i.e.,  the  greatest 
stress  of  the  same  kind  as  that  produced  by  the  dead  load 
occurs  when  the  moving  load  is  on  all  the  joints  between  the 
diagonal  in  question  and  the  farther  abutment ;  whereas  the 
greatest  stress  of  the  opposite  kind  occurs  when  the  moving 
load  covers  all  the  joints  between  the  diagonal  in  question  and 
the  nearer  abutment. 

The  work  of  determining  the  greatest  shearing-forces  may 
be  arranged  as  in  tables  on  p.  191. 

Counterbraces.  —  If  the  truss  were  constructed  with  those 
diagonals  only  that  slope  downwards  towards  the  centre,  and 
which  may  be  called  the  main  braces,  the  diagonals  11—12, 
13-14,  14-17,  and  16-19  would  sometimes  be  called  upon  to 
bear  a  thrust,  and  the  verticals  12-13  and  17-16  a  pull:  this 
would  necessitate  making  these  diagonals  sufficiently  strong 
to  resist  the  greatest  thrust  to  which  they  are  liable,  and  fixing 
the  verticals  in  such  a  way  as  to  enable  them  to 'bear  a  pull. 

In  order  to  avoid  this,  the  diagonals  10-13,  12-15,  15-16, 
and  17-18  are  inserted,  which  are  called  counterbraces,  and 
which  come  into  action  only  when  the  corresponding  main 


TRUSSES    WITH   VERTICAL   AND   DIAGONAL   BRACING.  2OI 


braces   would   otherwise   be   subjected   to   thrust.     They  also 
prevent  any  tension  in  the  verticals. 


Diagonals. 

Greatest  Shearing-Forces  of  the  Same  Kind 
Dead  Load. 

as  those  produced  by 

I-   2       28-26 

3^(1+2  +  3  + 

...+I3) 

=  227500 

3-  4     27-24 

3-£p(i  +  2+3+ 

...  +  I2)-^(I) 

=  I94M3 

5-  6     25-22 

3S2?(l  +  2  +  3  + 

J4 

=  162429 

7-  8     23-20 

^f(l  +  2  +  3  + 

.  ..+10)-  ^(1  +  2- 

f  3)           =  132357 

9-10     21-18 

^(1  +  2+3+ 

•••+  9)-I-^?(I  +  2- 

f  ...  +4)  =  103929 

11-12     19-16 

^(1  +  2+3+ 

...+  8)-^p(i  +  2- 

f...+5)=    77M3 

13-14     17-14 

(l  -J~  2-f-  3~t~ 
M 

•••+  7)-^i  +  2- 

f  ...+6)=    52000 

Diagonals. 

Greatest  Shearing-Forces  of  the  Opposite  Kind  to  those  produced  by 
Dead  Load. 

I3-H 

17-14 

3^?(i  +  2+3+  .  •  •  +  6)—  "—(i  +  2+.  .  .+7)  =     28500 

11-12 

I9-I6 

35000X                                                              v            12000,                                           ov                      f-f- 

^4                                                            ^4                      •  i 

9-10 

2I-I8 

^(,  +  2+        ...  +  4)--^(I  +  2+...+9)  =  -I357r 

The  main  braces  and  counterbraces  of  a  panel  are  never  in 
action  simultaneously.  Hence  we  have,  for  the  greatest  stresses 
in  the  diagonals,  the  following  results,  obtained  by  multiplying 

the  corresponding  shearing-forces  by 


cos  45' 


—  1.414. 


2O2 


APPLIED   MECHANICS. 


In  the  following  I  have  used  this  number  to  three  decimal 
places,  as  being  sufficiently  accurate  for  practical  purposes. 


Stresses  in  Main  Braces. 

Stresses  in  Counterbraces. 

I-    2 

28-26 

-321685 

15-12 

15-16 

-40299 

3-  4 

27-24 

-274518 

I3-IO 

I7-I8 

-  9393 

5-6 

25-22 

-229675 

7-8 

23-20 

-187153 

9-10 

21-18 

—  146956 

11-12 

19-16 

—  109080 

I3-M 

17-14 

-  73528 

Vertical  Posts.  —  Since  the  loads  are  applied  at  the  upper 
joints,  the  conditions  of  equilibrium  at  the  lower  joints  require 
that  the  thrust  in  any  vertical  post  shall  be  equal  to  the  vertical 
component  of  the  tension  in  that  diagonal  which,  being  in  action 
at  the  time,  meets  it  at  its  lower  end. 

Hence  it  is  equal  to  the  shearing-force  in  that  panel  where 
the  acting  diagonal  meets  it  at  its  lower  end. 

We  therefore  have,  for  the  posts,  the  following  as  the  greatest 
thrusts :  — 

STRESSES   IN   VERTICALS. 


3-  2 

27-26 

-J-2275OO 

5-  4 

25-24 

+  I94I43 

7-  6 

23-22 

+  162429 

9-  8 

2I-2O 

+  I3*357 

II-IO 

19-18 

+  103929 

13-12 

17-16 

+   77M3 

i5-J4 

+  52000 

CONCENTRATING    THE  LOAD   AT  THE  JOINTS.          2OJ 


I 

m 

•o 

4k 

cj 

00 

£ 

B 

£ 

o 

£ 

8 

• 

R 

«o 

ea 

/ 

/ 

/ 

/ 

/ 

X 

X 

x 

x 

\ 

\ 

\. 

\ 

\ 

i 

I 

Fig.  115  shows  the  stresses  marked  on  the  diagram. 

§  154.  Manner  of  Concentrating  the  Load  at  the  Joints. 

—  In  using  the  methods  given  above,  we  are 

assuming  that  all  the  loads  are  concentrated 

at  the  joints,  and  that  none  are  distributed 

over  any  of  the  pieces.     As  far  as  the  mov- 
ing load  is  concerned,  and  also  all  of   the 

dead   load   except   the  weight  of   the  truss 

itself,  this  always  is,  or  ought  to  be,  effected ; 

and  it  is  accomplished  in  a  manner  similar 

to  that  adopted  in  the  case  of  roof-trusses. 

This  method  is   shown  in  the  figure   (Fig. 

1 1 6);  floor-beams  being  laid  across  from 
girder  to  girder  at  the  joints, 
on  top  of  which  are  laid  longi- 
tudinal beams,  and  on  these 
the  sleepers  if  it  is  a  railroad 
bridge,  or  the  floor  if  it  is  a 
road  bridge.  The  weight  of 
the  truss  itself  is  so  small  a 
part  of  what  the  bridge  is 
called  upon  to  bear,  that  it 
can,  without  appreciable  error, 
be  considered  as  concentrated 
at  the  joints  either  of  the  up- 
per chord,  of  the  lower  chord, 
or  of  both,  according  to  the 
manner  in  which  the  rest  of 
the  load  is  distributed. 

§  155.  Closer  Approxima- 
tion to  Actual  Shearing- 
Force.  —  In  our  computations 
of  greatest  shearing-force,  we 


FIG.  116. 


\ 

FIG.  115. 


make  an  approximation  which  is  generally  considered  to  be 


2O4  APPLIED   MECHANICS. 

sufficiently  close,  and  which  is  always  on  the  safe  side.  To 
illustrate  it,  take  the  case  of  panel  3-5  of  the  last  example. 
In  determining  its  greatest  shearing-force,  we  considered  a  load 
of  35000  Ibs.  per  panel  point  to  rest  on  all  the  joints  from  the 
right-hand  support  to  joint  5,  inclusive,  and  the  dead  load  to 
rest  on  all  the  other  joints  of  the  truss.  Now,  it  is  impossible, 
if  the  load  is  distributed  uniformly  on  the  floor  of  the  bridge, 
to  have  a  load  of  35000  Ibs.  on  5  and  12000  on  3  simultaneously  ; 
for,  if  the  moving  load  extended  on  the  bridge  floor  only  up  to 
5,  the  load  on  5  would  be  only  12000  +  J(23°°°)  —  23500  Ibs., 
and  that  on  3  would  then  be  12000  Ibs.  If,  on  the  other  hand, 
the  moving  load  extends  beyond  5  at  all,  as  it  must  if  the  load 
on  5  is  to  be  greater  than  23500  Ibs.,  then  part  of  it  will  rest 
on  3,  and  the  load  on  3  will  then  be  greater  than  12000  Ibs.  ; 
for  whatever  load  there  is  between  3  and  5  is  supported  at 
3  and  5. 

Moreover,  we  know  that  the  effect  of  increasing  the  load  on 
5  is  to  increase  the  shearing-force,  provided  we  do  not  at  the 
same  time  increase  that  on  3  so  much  as  to  destroy  the  effect 
of  increasing  that  on  5- 

Hence,  there  must  be  some  point  between  3  and  5  to  which 
the  moving  load  must  extend  in  order  to  render  the  shearing- 
force  in  panel  3-5  a  maximum. 

Let  the  distance  of  this  point  from  5  be  x  ;  then,  if  we  let 

w  —  =  moving  load  per  foot  of  length, 

Moving  load  on  panel  =  wx, 

Part  supported  at  3      =  —  , 

20 


Part  supported  at  5       =  wx 

20 


Hence,  portion  of  shearing-force  due  to  the  moving  load  on 
panel  3-5  equals 


/ 


CONCENTRATING    THE  LOAD   AT  THE  JOINTS. 


12  /  WX2\  I     WXZ          W  I  \*\X*\ 

—  (  WX   --  -  )   ---   =   —  (  I2X    --  ^—  ). 
I4\  20  /  14     20  I4\  20    / 

This  becomes  a  maximum  when  its  first  differential  co-efficient 
becomes  zero,  i.e.,  when 

12  -  {%x  =  o; 
therefore 

x    =  9^.23. 

Hence,  when  the  moving  load  extends  to  a  distance  of  9.23  feet 
from  5,  then  the  shearing-force  in  panel  3-5,  and  hence  the 
stress  in  diagonal  3-4,  is  a  maximum. 


Panels. 

Portion  of  Shearing-Force 
due  to  Moving  Load  on 
Panel. 

Value 
of  xt  in 
feet. 

Portion  of  Load 
at  Joints  named 
below. 

Portion  of  Load 
at  Joints  named 
below. 

I-  3 

27-28 

^(.3*  -  ^ 

IO.OO 

I 

11500 

3 

11500 

I4\               20  / 

,, 

3-  5 

25-27 

,  ^/T0-t,       ^3^\ 

9-23 

3 

9797 

5 

11432 

\                      ) 

14  \               20  / 

5-  7 

23-25 

^Yii*  _  ISf!} 

8.46 

5 

8230 

7 

11227 

I4\               20  / 

7-  9 

21-23 

.^Yio*  -  ^\ 

7.69 

7 

6801 

9 

10886 

I4\               20  / 

9-1  1 

19-21 

w/              I3*2\ 

6.92 

9 

5507 

ii 

10409 

\       j'*'                                I 

I4\               20  / 

11-13 

17-19 

w  (  9.v        I3jr2\ 

6.15 

ii 

4350 

13 

9795 

\    *^                        y 

14  v             20  / 

/                                              2\ 

I3-I5 

15-17 

g  ?"  -  ^ 

5.38 

13 

3329 

15 

9045 

To  show  how  the  adoption  of  this  method  would  affect  the 
resulting  stresses  in  the  diagonals  and  verticals,  I  have  given 
the  work  above,  and  shown  the  difference  between  these  and 


2O6 


APPLIED   MECHANICS. 


the  former  results.     In  this  table  x  =  distance  covered  by  load 
from  end  of  panel  nearest  the  centre. 


Panels. 


Greatest  Shearing-Force  of  Same  Kind  as  that  due  to  Dead  Load. 


!-   3 

3-  5 
5-  7 
7-  9 


11-1 


27-28 

25-27 

23-25 
21-23 
19-21 
17-19 
I5-I.7 


(i  +...+«  3) 


35OO° 


35000 


(I+... 


^(i+...+  6)+£ 
14  v         14 


=«75°o 
=193385 


=130461 
=101654 
)=  74616 
1=  49345 


Hence,  for  stresses  in  main  braces,  we  have 


Diagonals. 

Stresses. 

I-    2 

28-26 

-321685 

3-  4 

27-24 

—  273446 

5-  6 

25-22 

—  227708 

7-  8 

23-20 

—  184472 

9-10 

21-18 

-143739 

11-12 

19-16 

-105507 

IS-M 

17-14 

—    69774 

Moreover,  for  the  shearing-forces   of   opposite   kind   from 


CONCENTRATING    THE   LOAD   AT  THE  JOINTS.          2O/ 


those  due  to  dead  load,  we  have,  if  x  =  distance  from  end  of 
panel  nearest  support  which  is  covered  by  moving  load,  — 


Panels. 

Portion  of  Shear  due 
to  Moving  Load  on  Panel. 

Value 
of  jr. 

Portion  of  Load 
at  Joints  named 
below. 

Portion  of  Load 
at  Joints  named 
below. 

^s 

17-15 

W/^               IT>X2\ 

1   fVK         _         ***     _     I 

I4\                   20  / 

4.62 

15 

2455 

13 

8171 

H-I3 

19-17 

™(sx  -  £1£!\ 

3-84 

13 

1695 

II 

7137 

I4\              20  / 

Panels. 


Greatest  Shearing-Forces  of  Opposite  Kind  from  those  due  to  Dead  Load. 


17-15 
19-17 


-  25846 


Hence  we  have  the  following  as  the  stresses  in  the  counter- 
braces  :  — 


Counter-Braces. 

Stresses. 

15-12 
13-10 

15-16 
I7-l8 

—    36546 

-       5820 

And,  for  the  verticals,  we  have  the  new,  instead  of  the  old, 
shearing-forces. 


208 


APPLIED  MECHANICS. 


The  following  table  compares  the  results  :  — 


\ 


Diagonals. 

Stress,  Ordinary 
Method. 

Stress,  New  Method. 

Difference. 

I—    2 

28-26 

-321685 

-321685 

-   3-  4 

27-24 

-274518 

-273446 

1072 

5-  6 

25-22 

-229675 

—  227708 

1967 

7-  8 

23-20 

-I87I53 

—  184472 

268l 

9-10 

21-18 

—  146956 

-M3739 

3217 

11-12 

19-16 

—  109080 

-105507 

3573 

iS-M 

17-14 

-   73528 

—   69774 

3754 

15-12 

15-16 

-  40299 

—  36546 

3753 

13-10 

17-18 

-     9393 

—   5820 

3573 

Verticals. 

Stress,  Ordinary 
Method. 

Stress,  New  Method. 

Difference. 

3-  2 

27-26 

+  227500 

+  227500 

O 

5-  4 

25-24 

+  I94I43 

+  193385 

758 

7-  6 

23-22 

+  162429 

+  161038 

J39i 

9-8 

2I-2O 

+  132357 

+  130461 

1896 

II-IO 

I9-I8 

+  103929 

+  101654 

2275 

13-12 

I7-I6 

+   77*43 

+    74616 

2527 

i5-J4 

+   28500 

+  49345 

2655 

§156.   Compound  Bridge-Trusses The  trusses  already 

discussed  have  contained  but  a  single  system  of  latticing,  or 


COMPO  UND   BRID  GE-  TR  USSES. 


209 


at  least  only  one  system  that  comes  in  play  at  one  time ;  so  that 
a  vertical  section  never  cuts  more  than  three  bars  that  are  in 
action  simultaneously,  the  main  brace  having  no  stress  upon  it 
when  the  counterbrace  is  in  action,  and  vice  versa. 

We  rrjay,  however,  have  bridge-trusses  with  more  than  one 
system  of  lattices ;  and,  in  determining  the  stresses  in  their 
members,  we  must  resolve  them  into  their  component  systems, 
and  determine  the  greatest  stress  in  each  system  separately, 
and  then,  for  bars  which  are  common  to  the  two  systems,  add 
together  the  stresses  brought  about  by  each. 

In  some  cases,  the  design  is  such  that  it  is  possible  to 
resolve  the  truss  into  systems  in  more  than  one  way,  and  then 
there  arises  an  uncertainty  as  to  which  course  the  stresses  will 
actually  pursue. 

In  such  cases,  the  only  safe  way  is  to  determine  the  greatest 
stress  in  each  piece  with  every  possible  mode  of  resolution  of 
the  systems,  and  then  to  design  each  piece  in  such  a  way  as  to 
be  able  to  resist  that  stress. 

Generally,  however,  such  ambiguity  is  an  indication  of  a 
waste  of  material ;  as  it  is  most  economical  to  put  in  the  bridge 
only  those  pieces  that  are  absolutely  necessary  to  bear  the 
stresses,  as  other  pieces  only  add  so  much  weight  to  the  struc- 
ture, and  are  useless  to  bear  the  load. 

The  mode  of  proceeding  can  be  best  explained  by  some 
examples. 

EXAMPLE  I.  —  Given  the  lattice-girder  shown  in  Fig.  117, 
loaded  at  the  lower  panel  points 

7^  1     3     5    7     9    11    13    15  17  19  21  23    25 

only.       Dead   load  =  7200  Ibs. 


per  panel    point,    moving   load 

=  18000  Ibs  per  panel   point; 

let  the  entire  length  of  bridge 

be  60  feet ;    let    the  angle   made    by  braces   with    horizontal 

=  60°. 


210 


APPLIED   MECHANICS. 


13        17 


23 


This  truss  evidently  consists  of  the  two  single  trusses  shown 
in  Figs,  nja 
and  I I jb ;  and 
we  can  compute 
the  greatest 
stress  of  each 


A/VVVYl 

8         12        16 


20        24 


FIG. 


FIG.  iiya. 

kind  in  each  member  of  these  trusses,  and  thus 
obtain  at  once 
all  the  diag- 
onal stresses, 
and  then,  by 
addition,  the 
greatest  chord  stresses. 

Thus  the  stress  in  1-3  (Fig.  117)  is  the 
same  as  the  stress  in  1-5  (Fig.  1170). 

The   stress   in    3-5  =  stress   in    1-5   (Fig. 
+  stress  in  3-7  (Fig.  117^). 

The    stress    in    5-7  =  stress    in    5-9    (Fig. 

7 a)  -p-  stress  in  3-7  (Fig.  117^).          V 

The  results  are  given  on  the  diagram  (Fig. 
the  work  being  left  for  the  student,  as 
it  is  similar  to  that  done  heretofore. 

EXAMPLE  II.  —  Given  the  lattice-girder 
shown  in  Fig.  1 1 8.  Given,  as  before,  Dead 
load  —  7200  Ibs.  per  panel  point,  moving  load 
=  18000  Ibs.  per  panel  point,  entire  length  of 
bridge  =  25  feet ;  load  applied  at  lower  panel 
points. 


Solution.  —  In  this  case,  there  are  two  possible  modes  of 
resolving  it  into  systems.  The  first  is  shown  in  Figs.  uSa  and 
1 1 8£:  and  this  is  necessarily  the  mode  of  division  that  must 
hold  whenever  the  load  is  unevenly  distributed,  or  when  the 


COMPOUND   BRIDGE-TRUSSES. 


211 


travelling-load  covers  only  a  part  of  the  bridge ;  for  a  single 
load  at  6  is  necessarily  put  in  communication  with  the  support 
at  2  by  means  of  the  diagonals  6-3  and  3-2,  and  with  the  sup- 
port at  12  by  means  of  the  diagonals  6-7,  7-10,  10-11,  and  the 
vertical  11-12,  and  can  cause  no  stress  in  the  other  diagonals. 


1357 


2     4      C      8     10    12 
FIG.  118. 


6  10    12 

FIG.  n8a. 


24  8  12 

FIG.  1 1 8£. 


2 .11 


357 


10    12 


FIG.  n8<r. 


When,  however,  the  whole  travelling-load  is  on  the  bridge, 
it  is  perfectly  possible  to  divide  it  into  the  two  trusses  shown 
in  Figs.  n&:  and  ii&/,  the  diagonals  4-5,  7-10,  6-7,  and  5-8 
having  no  stress  upon  them. 

When  the  load  is  unevenly  distributed,  we  have  certainly 
the  first  method  of  division  ;  and  when  evenly,  we  are  not  sure 
which  will  hold. 

Hence  we  must  compute  the  greatest  stresses  with  each 
mode  of  division,  and  use  for  each  member  the  greatest ;  for 
thus  only  shall  we  be  sure  that  the  truss  is  made  strong 
enough. 

We  shall  thus  have  the  following  results  :  — 


212 


APPLIED   MECHANICS. 


FIRST   MODE  OF   DIVISION    (FIGS.  n8a   AND 


Diagonals. 

Greatest  Shearing-Force 
of  One  Kind. 

Greatest  Shearing-Force 
of  Opposite  Kind. 

Corresponding 
Stresses. 

Fig. 

Fig. 

2-  3 

12-9 

—  (3  +  i)       =  2Ol6o 

0 

+23279 

—            O 

3-6 

9-3 

^^(3  +  i)      =  20160 
5 

0 

-23279 

+       ° 

6-7 

8-5 

^_z^?(2)=  2l6o 

»Jp(2)_Z2p  =    8640 

+   2494 

—  9976 

7-10 

5-4 

25200           72OO, 

(2)  =     2160 

5               5 

—  (2)  =    8640 

—    2494 

+  9976 

IO-II 

4-1 

0                              =             0 

25200, 
•+--(2  +  4)      =  30240 

0 

-34918 

Chords. 

Supporting  force  at  2  (Fig.  1180)  or  12  (Fig. 

=  ^(3  +  1)  =  20160, 

Supporting  force  at  12  (Fig.  1180)  or  2  (Fig.  ii8£) 


Section. 

Chords. 

Maxi- 

!    mum 

Corn- 

Bending-Moment. 

;  Stresses 
in 

Chords.   !   p°nenls 

of 

Greatest 
Resultant 

3 

gi 

«      !    ~* 

jj 

oo 

H      Separate 

Stresses. 

Stresses. 

.SP 

.£? 

be 

fci    j  Trusses. 

b 

£ 

:     b 

g< 

3 

9 

201  60  X    5  =  100800 

2-  6 

8-12 

-11639 

1-3 

; 
9~II          0+1-5 

+  17459 

6 

8 

20l6oXlO  =  20l6oO 

3-  7 

5-  9 

+23279 

3-5 

7-  9,3-  7+1-5 

+40738 

7 

5 

20160X15  —  25200 

X  5  =  176400 

6-10 

4-  8 

—20369 

5-7 

J3-  7+5-9 

+46558 

10 

4 

30240X  5  =  151200 

7-1  1 

i-  5 

+  17459 

2-4 

10-12  2-   6+2-4 

-11639 

10-12 

2-  4 

O 

4-6 

8-10,2-  6+4-8 

-32008  i 

6-8 

|6-ro+4-8|  -40738  ; 

COMPOUND   BRIDGE-TRUSSES. 


213 


SECOND    METHOD    OF    DIVISION    (FIGS.  nSc    AND    1180?). 

Diagonals  (Fig. 


Diagonals. 

Maximum 
Shear. 

Corresponding 
Stresses. 

1-4 

IO-II 

252OO 

—  29098 

4-5 

7-10 

O 

0 

Fig.  u8</. 


Diagonals. 

Maximum 
Shear. 

Corresponding 
Stresses. 

2-3 

9-12 

252OO 

+  29098 

3-6 

8-  9 

25200 

—  29098 

6-7 

5-8 

0 

O 

Chords. 

Each  supporting  force  in  either  figure  =  25200. 
Fig.  1 1 8*. 

Bending-moment  anywhere  between  4  and  10  =  (25200)  (5)  =  126000; 

/.     Stress  in  i-u  =  +14549, 
.-.     Stress  in  4-10  =  —14549. 

Fig.  nSd. 

Bending-moment  at  3  or  9  =  126000, 

Bending-moment  anywhere  between  6  and  8  =  252000; 

/.     Stress  in  3-9  =  +29098, 

Stress  in  2-6  or  8-12  =  —14549, 
Stress  in  6-8  =  —29098. 


214 


APPLIED   MECHANICS. 


Hence  we  have  for  chord  stresses,  with  this  second  divis- 
ion, — 


Chords. 

Stresses. 

1-3 

9-1  1 

I-II    +  0 

+  14549 

3-5 

7-  9 

i-n  +  3-9 

+  43647 

5-7 

- 

i-n  +  3-9 

+  43647 

2-4 

IO-I2 

0   -+-    2-6 

-14549 

4-6 

8-10 

4-10  -f  2-6 

—  29098 

6-8 

— 

4-10  +  6-8 

-43647 

Hence,  selecting  for  each  bar  the  greatest,  we  shall  have,  as 
the  stresses  which  the  truss  must  be  able  to  resist,  — 


1-4 

10-11 

+•        o 

-34918 

J-3 

9-1  1 

+  17459 

2-3 

12-  9 

+  29098 

o 

3-5 

7-  9 

+43647 

3-6 

9-  8 

+         o 

—  29098 

5-7 

- 

+46558 

4-5 

10-  7 

+  9976 

—  2494 

2-4 

IO-I2 

—  M549 

5-3 

7-6 

+  2494 

-  9976 

4-6 

8-10 

—  32008 

6-8 

—43647 

These  results  are  recorded  in  Fig.  uSe. 

( D+17549  (3)+  43C47(5)+  46558(7)+ 43647(9)+ 17459(11) 


'  (2  WI4549<4)-  32008  (6)  -  43647  (8)-32008  C10)-14549a2) 
FIG.  n&r. 


§157.  Other  Trusses.  —  In    Figs.  119,    120,   and    121,  we 
have  examples  of  the  double-panel  system  with  the  load  placed 


OTHER    TRUSSES. 


215 


at  the  lower  panel  points  only.  When,  as  in  119  and  120,  the 
number  of  panels  is  odd,  the  same  ambiguity  arises  as  took  place 
in  Fig.  1 1 8.  When,  on  the  other  hand,  the  number  of  panels 
is  even,  as  shown  in  Fig.  121,  there  is  only  one  mode  of  division 
into  systems  possible.  The  diagrams  speak  for  themselves,  and 
need  no  explanation. 


2   4    6   8   10   12  14   16   18   20   22   24  26   28   80   82 


9   11   13   15   17   19   21   23   25  27   29   31    33   34 


185 


2   4 


12      16 


24      28 


13      17 

FIG.  uga. 


25      29       33  34 


10      14      18      22       26      30 


7       11       15       19      23       27      31   38   34 
FIG.  ngJ. 


2    4 


22 26 30 


17   19      23 
FIG.  ngc. 


31   33   34 


24      28       32 


11       15          21       25      29       33 
FIG.  ngd. 


216 


APPLIED   MECHANICS. 


246        8       10       12       14       16      18      20       22      24      26      28       30      32      34      36 


3        5       7        9       11       13       15      17      19       21      23      25      27      29      31       33      35 


24  8  12  16 


24  28  32 


13  7 


11  15  19 

FIG.  i2oa. 


27  31  35 


2 6 10  14  18  22  2«  30  34      36 


159 


13  17 

FIG. 


21  25  29  33      35 


2  6  10  14 


24  28  32  36 


1  5  9 


13  17      19 

FlG.  T.2QC. 


23  27  31  35 


Z4 36 


13  7  11     "          15  21  25  29  33       35 

FIG.  i2od. 


2        4        6        8       10       12       14      16       18 


X 


X 


\ 


3        5        7        9       11       13      15     17        19 
FIG.  120^. 


FINK'S   TRUSS. 


217 


2 6 10 U 18 


1  11  15 

FIG.  i2i«. 


20 


2        4 


12 16       18 


V 

/        N 


5  9  13 

FIG.  i2i3. 


20 


FIG.  122. 


The  trusses  given  above  may  be  considered  as  examples,  to 
be  solved  by  the  student  by  assuming  the  dead  and  the  moving 
load  per  panel  point  respectively. 

§158.  Fink's  Truss.  —  The  description  of   this  truss  will 
be  evident  from  the  figure.     There  is,  first,  the  primary  truss 
i -8- 1 6  ;  then  on  each  side 
of  9-8  (the  middle  post  of 
this    truss)    is  a  secondary 
truss    (1-4-9   on    tne    1^ 
and  9-12-16  on  the  right). 

Each  of  these  secondary 

trusses  contains  a  pair  of  smaller  secondary  trusses,  and  the 
division  might  be  continued  if  the  segments  into  which  the 
upper  chord  is  thus  divided  were  too  long. 

Of  the  inclined  ties,  there  is  none  in  which  any  load  tends 
to  produce  compression  ;  in  other  words,  every  load  either  in- 
creases the  tension  in  the  tie,  or  else  does  not  affect  it.  Hence 


218 


APPLIED   MECHANICS. 


the  greatest  stresses  in  all  the  members  will  be  attained  when 
the  entire  travelling-load  is  on  the  truss,  and  we  need  only  con- 
sider that  case. 

The  determination  of  the  stress  in  any  one  member  can 
readily  be  obtained  by  determining,  by  means  of  the  triangle 
of  forces,  the  stress  in  that  member  due  to  the  presence  of 
the  total  load  per  panel  point,  at  each  point,  and  then  adding  the 
results.  This  will  be  illustrated  by  a  few  diagonals. 

Let  angle  8-1-9  —  ^ 
Let  angle  4-1-5  —  ilt 
Let  angle  2-1-3  =  **  J 

we  shall  have,  if  w  +  wt  —  entire  load  per  panel  point,  — 


Designation 
of  Ties. 

EFFECT  OF  LOADS  AT 

Resultant 
Tensions. 

3 

6 

1 

9 

11 

13 

15 

1-2 

2-5 
5-6 
6-9 
1-4 

4-9 
i-8 

w  +  Wi 

0 
0 

o 

0 
W  +  Wi 

0 

0 
W  +  Wi 

0 
0 

o 

0 

o 

0 
W  +  M>i 

0 
0 

o 
o 

0 

o 
3  w+wi 

0 
O 

o 

0 

o 

0 
W  +  Wi 

0 

o 
o 

0 

o 

0 
W  +  Wi 

W  +  Wi 

2  sin  z'2 

IV  +  Wi 

2  sin  z'2 

2#  +  Wi 

2  sin  z2 
o 

0 
W  +  Wi 

2  sin  /2 
w  -\-  Wi 

2  sm  /2 
w  -f-  Wi 

2  sin  ?2 
w  +  wl 

2  sin  /2 

W  +  Wi 

2  sin/2 
ay  +  Wi 

4  sin  *, 

W  +  Wi 

2  sin  *, 
w  +  Wi 

4  sin  z'x 
w  -f-  Wi 

sin  /! 
w  +  Wi 

4  sin  ii 

W  +  Wi 

2  sin  i-i 

W  +  Wi 

4  sin  z'j 
3  w-\-Wi 

sin  *! 
2(w-j-Wl) 
sin  i 

8  sin  i 

4  sin  i 

8    sin  e 

2  sin  * 

8    sin  e 

4  sin  /' 

8  sin  i 

The  stresses  in  all  the  other  members  may  be  found  in  a 
similar  manner. 


GENERAL   REMARKS. 


2I9 


§  159.  Bollman's  Truss.  —  The  description  of  this  truss  is 
made  sufficiently  clear  by  the  figure.  The  upper  chord  is  made 
in  separate  pieces  ;  and 

1  3  5  7  9  11  12 

the  short  diagonals  2-5, 
3-4,  4-7,  5-6,  7-8,  6-9, 
8-1 1,  and  9-10  are  only 
needed  to  prevent  a 
bending  of  the  upper 
chord  at  the  joints. 
When  this  is  their  only  object,  the  stress  upon  them  cannot  be 
calculated  :  indeed,  it  is  zero  until  bending  takes  place ;  and 
then  it  is  the  less,  the  less  the  bending.  Hence,  in  this  case, 
the  stress  is  wholly  taken  up  by  the  principal  ties ;  and  these 
have  their  greatest  stress  when  the  whole  load  is  on  the  bridge. 
The  computation  of  the  stresses  is  made  in  a  similar  man- 
ner to  that  used  in  the  Fink. 

§  1 60.  General  Remarks.  —  The  methods  already  explained 
are  intended  to  enable  the  student  to  solve  any  case  of  a  bridge- 
truss  where  there  is  no  ambiguity  as  to  the  course  pursued  by 
the  stresses. 

In  cases  where  a  large  number  of  trusses  of  one  given  type 
are  to  be  computed,  it  would,  as  a  rule,  be  a  saving  of  labor  to 
determine  formulae  for  the  stresses  in  the  members,  and  then 
substitute  in  these  formulae. 

Such  formulae  may  be  deduced  by  using  letters  to  denote 
the  load  and  dimensions,  instead  of  inserting  directly  their 
numerical  values ;  and  then,  having  deduced  the  formulae  for 
the  type  of  truss,  we  can  apply  it  to  any  case  by  merely  sub- 
stituting for  the  letters  their  numerical  values  corresponding 
to  that  case. 

Such  sets  of  formulae  would  apply  merely  to  specific  styles 
of  trusses,  and  any  variation  in  these  styles  would  require  the 
formulae  to  be  changed. 


22O  APPLIED  MECHANICS. 

In  order  to  show  how  such  formulae  are  deduced,  a  few  will 
be  deduced  for  such  a  bridge  as  is  shown  in  Fig.  in. 

Let  the  load  be  applied  at  the  upper  panel  points  only  ;  let 
dead  load  per  panel  point  =  w,  moving  load  per  panel  point 
=  wt.  Let  the  whole  number  of  panels  be  N,  N  being  an  even 
number.  Let  the  length  of  one  panel  =  height  of  truss  =  /. 
Then  length  of  entire  span  =  Nl. 

Consider  the  (n  +  i)th  panel  from  the  middle. 

The  stress  in  the  main  tie  is  greatest  when  the  moving  load 
is  on  all  the  panel  points  from  the  farther  abutment  up  to  the 
panel  in  question,  (n  +  i)th. 

Hence,  for  the  »th  panel  from  the  middle,  the  greatest  shear- 
ing-force that  causes  tension  in  the  main  tie  is  equal  to 


Hence  stress  in  main  tie 


For  the  counterbrace,  we  should  obtain,  in  a  similar  way,  the 
formula 


N  . 

w.\ n\ \- 

J  2 


n     —  wN(2n  -f-  i) 


which  represents  tension  when  it  is  positive.     Proceed   in   a 
similar  way  for  the  other  members. 

When  there  is  more  than  one  system,  we  must  divide  the 
truss  into  its  component  systems;  and  when  there  is  ambiguity, 
we  must  use,  in  determining  the  dimensions  of  each  member, 
the  greatest  stress  that  can  possibly  come  upon  it. 


CENTRE   OF  GRAVITY.  221 


CHAPTER  V. 

% 
CENTRE    OF   GRAVITY. 

§  1  6  1.  The  centre  of  gravity  of  a  body  or  system  of  bodies,  is 
that  point  through  which  the  resultant  of  the  system  of  parallel 
forces  that  constitutes  the  weight  of  the  body  or  system  of 
bodies  always  passes,  whatever  Be  the  position  in  which  the 
body  is  placed  with  reference  to  the  direction  of  the  forces. 

§  162.  Centre  of  Gravity  of  a  System  of  Bodies.  —  If 
we  have  a  system  of  bodies  whose  weights  are  Wlf  W2,  Wy  etc., 
the  co-ordinates  of  their  individual  centres  of  gravity  being 
(*»  y»  -O,  (*2,  y«  z2),  (xy  yy  #3),  etc.,  respectively,  and  if  we 
denote  by  xQy  y0,  zm  the  co-ordinates  of  the  centre  of  gravity  of 
the  system,  we  should  obtain,  just  as  in  the  determination  of  the 
centre  of  any  system  of  parallel  forces,  — 

i°.  By  turning  all  the  forces  parallel  to  OZ>  and  taking 
moments  about  OYy 


(W,  +  W2  +  W3  +  etc.)*0  =   WiXl  +  W2x2  +  0>3  +  etc., 
or 


and,  taking  moments  about  OX, 

W2y2  +  W,y,  +  etc., 


or 


222 


APPLIED   MECHANICS. 


2°.  By  turning  all   the  forces  parallel  to  OX,  and  taking 
moments  about  OY, 


,  4- 


etc.)*,  =  Wjt 


etc., 


or 


Hence  we  have,  for  the  co-ordinates  of  the  centre  of  gravity 
of  the  system, 


Jo  = 


EXAMPLES. 

i.   Suppose  a  rectangular,  homogeneous  plate  of  brass  (Fig.  125), 

where  AD  —  1 2  inches,  AB  =  5  inches, 
and  whose  weight  is  2  Ibs.,  to  have 
weights  attached  at  the  points  A,  B,  C, 
and  D  respectively,  equal  to  8,  6,  5,  and 
— x  ^  Ibs. ;  find  the  centre  of  gravity  of  the 
system. 


Solution. 
Assume  the  origin  of  co-ordinates  at 


the  centre  of  the  rectangle,  and  we  have 

W,  =  2,      W2  =  8,     W,  =      6,     W,  =  5,     Ws  =      3, 

Xt    =  o,      x2    =6,      x3    =       6,      x4    =  —  6,      xs    =  —  6, 

j'l   =  o.    j'a  =  i    J3    =  -i    .>'4  =  -i    y,  =     f  ; 

=  o  +  48  +  36  —  30.0  —  18.0  =  36, 

=  o  +  20  —  15  —  12.5  -h  7.5  =_   o, 

=  2  -f-     8  H-     6+     5.0+  3.0=24; 

36  o 


Hence  the  centre  of  gravity  is  situated  at  a  point  E  on  the  line  OX, 
where  (y.£  =  1.5. 


CENTRE   OF  GRAVITY  OF  HOMOGENEOUS  BODIES.     22$ 

2.  Given  a  uniform  circular  plate  of  radius  8,  and  weight  3  Ibs, 
(Fig.  126).  At  the  points  A,  B,  C,  and  D, 
weights  are  attached  equal  to  10,  15,  25,  and  23 
Ibs.  respectively,  also  given  AB  =  45°,  BC  = 
105°,  CD  —  120°  \  find  the  centre  of  gravity  of 
the  system.  • 


§  163.  Centre  of  Gravity  of  Homogeneous  Bodies.  —  For 
the  case  of  a  single  homogeneous  body,  the  formulae  have  been 
already  deduced  in  §  44.  They  are 

fxdV  fydV  fzdV 

~ 


and  for  the  weight  of  the  body,   . 

W  =  wfdV, 

where  xm  j0,  zm  are  the  co-ordinates  of  the  centre  of  gravity  of 
the  body,  W  its  weight,  and  w  its  weight  per  unit  of  volume. 

From  these  formulae  we  can  readily  deduce  those  for  any 
special  cases  ;  thus,  — 

(a)  For  a  volume  referred  to  rectangular  co-ordinate  axes, 
dV  '=  dxdydz. 

_  fffxdxdydz          __  fffydxdydz  _  fffzdxdydz 

fffdxdydz'    y°    =   fff  dxdydz'     *°  ==  fffdxdydz' 

(b)  For  a  flat  plate  of  uniform  thickness,  /,  the  centre  of  grav- 
ity is  in  the  middle  layer  ;    hence   only  two  co-ordinates  are 
required  to  determine  it.     If  it  be  referred  to  a  system  of  rect- 
angular axes  in  the  middle  plane,  dV  '  =  tdxdyy 

ffxdxdy  _  ffydxdy 

ffdxdy'    y°~  ffdxdy' 


224  APPLIED   MECHANICS. 

The  centre  of  gravity  of  such  a  thin  plate  is  also  called  the 
centre  of  gravity  of  the  plane  area  that  constitutes  the  middle 
plane  section  ;  hence  — 

(c)  For  a  plane  area  referred  to  rectangular  co-ordinate  axes 
in  its  own  plane, 

_  ffxdxdy  _  ffydxdy 

~~ 


(d}  For  a  slender  rod  of  itniform  sectional  area,  a,  if  x,  y,  zy 
be  the  co-ordinates  of  points  on  the  axis  (straight  or  curved)  of 
the  rod,  we  shall  have  dV  =  ads  =  a^l  (dx)2  +  (dy)2  +  (dz}2, 


fds 


_  /*& 

*o     —   •    .,  , 


(e)  For  a  slender  rod  whose  axis  lies  wholly  in  one  plane, 
the  centre  of  gravity  lies,  of  course,  in  the  same  plane ;  and  if 
our  co-ordinate  axes  be  taken  in  this  plane,  we  shall  have  z  =  o 

-T-  =  o,  and  also  z0  =  o.     Hence  we  need  only  two  co- 


CENTRE   OF  GRAVITY  OF  HOMOGENEOUS  BODIES.      22$ 

ordinates  to  determine  the  centre  of  gravity,  hence  dV '=  ads. 


>24- 

_Jxds 
*°     :  Sds 


(/)  For  a  line,  straight  or  curved,  which  lies  entirely  in  one 
plane,  we  shall  have,  again, 


Sds 


Whenever  the  body  of  which  we  wish  to  determine  the 
centre  of  gravity  is  made  up  of  simple  figures,  of  which  we 
already  know  the  positions  of  the  centres  of  gravity,  the  method 
explained  in  §  162  should  be  used,  and  not  the  formulae  that 
involve  integration  ;  i.e.,  taking  moments  about  any  given  line 
will  give  us  the  perpendicular  distance  of  the  centre  of  gravity 
from  that  line. 

In  the  case  of  the  determination  of  the  strength  and  stiff- 
ness of  beams,  it  is  necessary  to  know  the  distance  of  a  hori- 
zontal line  passing  through  the  centre  of  gravity  of  the  section, 


226  APPLIED  MECHANICS. 

from  the  top  or  the  bottom  of  the  section  ;  but  it  is  of  no  prac- 
tical importance  to  know  the  position  of  the  centre  of  gravity 
on  this  line.  In  most  of  the  examples  that  follow,  therefore, 
the  results  given  are  these  distances.  These  examples  should 
be  worked  out  by  the  student. 

In  the  case  of  wrought-iron  beams  of  various  sections,  on 
account  of  the  thinness  of  the  iron,  a  sufficiently  close  approxi- 
mation is  often  obtained  by  considering  the  cross-section  as 
composed  of  its  central  lines ;  the  area  of  any  given  portion 
being  found  by  multiplying  the  thickness  of  the  iron  by  the 
corresponding  length  of  line,  the  several  areas  being  assumed 
to  be  concentrated  in  single  lines. 

EXAMPLES. 

1 .  Straight  Line  AB  (Fig.  127).  —  The  centre  of  gravity  is  evidently 

at  the  middle  of  the  line,  as  this  is  a  point  of 

A  C  g 

1 •     symmetry. 

FIG.  127. 

2.  Combination  of  Two  Straight  Lines.  —  The  centre  of  gravity  in 
each  case  lies  on  the  line  OOl}  Figs.  128,  129,  130,  and  131. 

(a)  Angle-Iron  of  Unequal  Arms  (Fig.  128).  —  Length  AB  =  b, 
length  BC  =  h,  area  AB  =  A,  area 

BC  =  B; 

.0, 

bh  _^±_  J_A_ 

/.     BE  =  DE  =  \  .  A  DC 

\Jb2  +  h2  FIG.  128. 

(b)  Angle-Iron  of  Equal  Arms  (Fig.  129).  — Length  AB  =  BC 

B  =  b; 


taN*  BE  =  DE  =  ±_  =  ^ 


FIG.  129. 


CENTRE   OF  GRAVITY  OF  HOMOGENEOUS  BODIES.      22J 


(c)    Cross  of  Equal  Arms  (Fig.  130).  —  AB  =  OO,  =  h; 


.-.    AC  =  BC  =  -• 


c p, 


B 
FIG  130. 

d)  T-Iron  (Fig.  131).  —  Area  AB  =  A,   area  CE  =  B,  length 

A        E B  CE  =  h; 

Bh 


2(A  + 


FIG.  131. 


3.   Combination  of  Three  Lines. —  QO^  =  line  passing  through  the 
centre  of  gravity. 

(a)    Thin    Isosceles   Triangular  Cell  (Fig.  132).  —  Length 

BC  =  a,  length  AC  =  b,  area  AB  =  BC  A         D          c 

\ i          7 

ftj 


A  area  A  C  —  B  ;                                           o 

\ 

/ 

/.     DB  =  \  a*  -  - 
4 

\    |E 

\ 

FIG. 

/ 

132. 

4 

.        7~)/7                     """"             \!  (  f/j          h\  1 

>  +  «: 

-    2(2A    +    5)V(" 

4    1    j9 

7?/r  —                      I//O/T      ^^  ^ 

o/y   J_   7>\ 

6"^^^-  in  Different  Position  (Fig.  133). 


FIG.  133. 


228 


APPLIED   MECHANICS. 


(<r)  Channel-Iron  (Fig.  134).  —  Area  of  flanges  =  A,  area  of  web 
=  B,  depth  of  flanges  +  -J  thickness  of 
web  =  h; 

Ah 


CE 


B  c 

FIG.  134. 


^TTB' 


(d)  T-Beam   (Fig.  135).  —  Area  of  upper  flange  =  A^  area  of 

lower  flange  =  A2,  area  of  web  =  B,  height  =  h. 

A 
B 

_i_    /?'  G 


rr  -  h 
= 


2  A,  +  A2 


o, 


EOF 

FIG.  135. 


2  At  +  A2  +  B 

4.   Combination  of  Four  Lines. —  OOI  =  line  passing  through  the 
centre  of  gravity. 

(a)   Thin  Rectangular  Cell  (Fig.  136).  — Length  AB  =  h; 


-o, 


/.     AE  =  BE  =  - 


FIG.  136. 

(b)    Thin  Square  Cell  (Fig.  i$-]).  —  AB  = 


.-.     BE  =  CE  =  -• 


-o, 


FIG.  137. 

5.   Circular  Arcs. 

(a)    Circular  Arc  AB  (Fig.  138).  —  Angle  A  OB  =  0It  radius  =  r. 


Y  Use  formula 

A 


fxds 


/yds 

•* 


but  use  polar  co-ordinates,  where 
FIG.  138.  ds  =  rdO,     x  =  r  cos  0,    y  =  >  sin  0, 


CENTRE   OF  GRAVITY  OF  HOMOGENEOUS  BODIES.    '22Q 


POi 
2    I     COS 


t 

r2  I    sin  OdO 


.(sing,) 


(i  —  cos0t) 


Circular  Arc  AC  (same  figure). 
r  sin  0j 


Jo  = 


(<:)   Quarter- Arc  of  Circle  AB,  Radius  r  (Fig.  139), 


•I 


cos 


2r 


*=r 


(d)  Semi- circumference  ABC  (same  figure). 


FIG.  139. 


*o  =   — ,      Jo  =   O. 

7T 

6.   Combination  of  Circles  and  Straight  Lines. 

Barlow  Rail  (Fig.    140). — Two  quadrants,  radius    r,  and  web, 
c.        whose  area  =  -^  the  united  area  of  the  quadrants. 
">v  X*"J        Let  united  area  of  quadrants  =  A,  area  of  web 

/  let  EF  =  XQ  : 


=  -  =  EF,. 


FIG.  140. 


230 


APPLIED   MECHANICS. 


7.  Areas. 

(a)   T-Section  (Fig.  141).  —  Let  length  AB  —  B,  EF  =  b,  entire 
height  =  H,  GE  =  h.     Let  distance  of  centre 
of  gravity  below  AB  —  x^ ;   therefore,  taking 
1    moments  about  AB  as  an  axis, 

-  h(B  -  b)\ 


-  h(B  - 


whence  we  can  readily  derive  xt. 


(b)  I-Section  (Fig.  142).  — Let  AB  =  B,  GH  =  b,  MN  =  bl9 
entire  height  =  H,  BC  =  H  —  h,  EH  =  ht ;  and  let  xt  =  distance 
of  centre  of  gravity  below  AB.  A  B 

Hence,  taking  moments  about  AB,  vtQ  have 


l  \B(H  -  h) 


whence  we  can  deduce 


(r)   Triangle  (Fig.  143).  —  If  we  consider  the   triangle  OBC  as 
composed  of  an  indefinite  number  of  narrow  strips  parallel  to  the  side 
CB,  of  which  FLHK  is  one,  the  centre  of  gravity 
of  each  one  of  these  strips  will  be  on  the  line  OD 
drawn  from  O  to  the  middle  point   of  the  side 
CB  ;  hence  the  centre  of  gravity  of  the  entire  tri- 
D  c    angle  must  be  on  the  line  OD.     For  a  similar  rea- 

son, it  must  be  on  the  median  line  CE  ;  hence  the 
centre  of  gravity  must  be  at  the  intersection  of  the  median  lines,  and 
hence 


OG  =  \OD.     Moreover,  area  = 


BC .  ODsmODC 


CENTRE   OF  GRAVITY  OF  HOMOGENEOUS  BODIES.     231 


(a)    Trapezoid  (Fig.  144). 


First  Solution.—  Bisect  AB  in  O,  and  CE  in  D;  let  g^  be  the 
centre  of  gravity  of  CEB,  and  g^  that  of  ^^#C. 
Then  will  G,  the  centre  of  gravity  of  the  trape- 
zoid,  be  on  the  line  gjg^  and 


Ggl  _  ABC 
~  CEB' 


o  B 

FIG.  M4. 


But  it  must  be  on  the  line  OD;  hence  it  is  at  their  intersection. 
From  the  similarity  of  GGlgl  GG^gv  we  have 


GG,       Gg±      ABC      AB=  B . 

9  ~  Gg,  ~  BEC  ~~  CE~    b  ; 


GG,  b  •".  „  „       OD 

=        '  andsmce   G'G-=      f 


Second  Solution. — Fig.  144  (a).  Let  O  be  the 
point  of  intersection  of  the  non-parallel  sides  AC 
and  BE.  Let  OF  =  xlt  OD  =  *„  OG  =  #0.  Take 
moments  about  an  axis  through  O,  and  perpen- 
dicular to  OF)  and  we  readily  obtain 


A  F 

FIG. 


232  APPLIED   MECHANICS. 

(e)  Parabolic  Half-Segment   OAB    (Fig.    145).—  Let    OA  —  #„ 

AB  —  yl  ;  let  x0,yoy  be  the  co-ordinates  of  the  centre  of  gravity  ;  let 

the  equation  of  the  parabola  be  y2  = 


^    pzakxk  -^  g 

Jo     Jo      xdxdy       2a  J     x^dx 


FIG.  145. 


/jTj      Piatx* 
I      ydxdy 
*^°  i     i 


t/o  i 


/*i    r^i 
/     t 

2^^ 


/»*,   /»j»/1 
/      /     L 

1/0          ^ 

/^l 


J'l 


Area  =  JPI>;I  —  | 


Area 


(/)  Parabolic  Spandril  OBC  (Fig.  145).  —  Let  ^0,^0  be  co-ordi- 
nates of  centre  of  gravity  of  the  spandril. 


CENl'RE  OF  GA'A  VI TY  OF  HOMOGENEOUS  BODIES.       233 


(g]   Circular  Sector  OAC  (Fig.  146).— Let  OA  =  r,  AOX  =  #„ 
x0,jo,  be  the  co-ordinates  of  the  centre  of  gravity  : 


£. 

^  v 

*  +  /r 

t/0 

COS  0j        f*X 

tand, 

xdxdy 

I 

\ 

\ 

0 

/v    ~^-*dx 

_        r  COS 

7  j                                      _£" 

X 

uin  Vj 

/^     B 

x 

j 

-r(2rO^) 

O 

o  sin  ^. 

\L 

1 

• 

:^-^' 

xi/__, 

c 

•t       / 

yi   \ 

FIG.  146. 

Area  = 

Second  Solution. 

Consider  the  sector  to  be  made  up  of  an  indefinite  number  of 
narrow  rings  ;  let  p  be  the  variable  radius,  and  dp  the  thickness  : 

.*.     Elementary  area  =  2p&ldp, 

and  centre  of  gravity  of  this  elementary  area  is  on  OX,  at  a  distance 
/i 

from  O  equal  to  p — ^ — 1  [see  Example  5 


.'.     x  = 


/    2p0.dp  20.  /    pdp 

e/o  t/o 

(h)   Circular  Half -Segment  ABX  (Fig.  146), 


/r  ftVrt-j:*  (*r 

/         xdxdy  I     xVr*  - 

_  ,  oosg,^  °  __  _          **r  COS  flt 


__ 

Sector  minus  triangle 


x*dx 

—  %r~  sin  t^,  cos 
==*H 


in3^. 


sm 


,  —  sn    ,  cos 


/>r  /»r2  -  ^a 

Jr  cos  e  J°       y  '  _  i  4sin 

-  ^(0i  _  sin  ^  cos  ^)  ~  8 


i  4sina|^  —  sin'^cosfl, 
8,  -  sin  ^  cos  0, 


234  APPLIED   MECHANICS. 


§  164.  Pappus's  Theorems. — The  following  two  theorems 
serve  often  to  simplify  the  determination  of  the  centres  %of 
gravity  of  lines  and  areas.  They  are  as  follows : — 

THEOREM  I. — If  a  plane  curve  lies  wholly  on  one  side  of  a 
straight  line  in  its  own  plane,  and,  revolving  about  that  line, 
generates  thereby  a  surface  of  revolution,  the  area  of  the  sur- 
face its  equal  to  the  product  of  the  length  of  the  revolving  line, 
and  of  the  path  described  by  its  centre  of  gravity. 

Proof. — Let  the  curve  lie  in  the  xy  plane,  and  let  the  axis 
of  y  be  the  line  about  which  it  revolves.  We  have,  from  what 

precedes,  §  163  (e\  XQ  —  ~7^p 

.'.    x0fds  =  fxds9 

where  x0  equals  the  perpendicular  distance  of  the  centre  of 
gravity  of  the  curve  from  O  F,  ds  —  elementary  arc, 


or,  reversing  the  equation, 


But  f(2Ttx)ds  =  surface  described  in  one  revolution,  while  s  = 
length  of  arc,  and  2-rrx0  =  path  described  by  the  centre  of 
gravity  in  one  revolution.  Hence  follows  the  proposition. 

THEOREM  II.  —  If  a  plane  area  lying  wholly  on  the  same 
side  of  a  straight  line  in  its  own  plane  revolves  about  that  line, 
and  thereby  generates  a  solid  of  revolution,  the  volume  of  the 
solid  thus  generated  is  equal  to  the  product  of  the  revolving 
area,  and  of  the  path  described  by  the  centre  of  gravity  of  the 
plane  area  during  the  revolution. 


PAPPUS'S   THEOREMS.  235 

Proof.  —  Let  the  area  lie  in  the  xy  plane,  and  let  the  axis 
OY  be  the  axis  of  revolution.  We  then  have,  from  what  has 
preceded,  if  XQ  —  perpendicular  distance  of  the  centre  of  gravity 
of  the  plane  area  from  OY,  the  equation,  §  163  (b)t 

.  ffxdxdy 
X°  "   ffdxdy  ' 
Hence 

x0  ffdxdy          =  ffxdxdy; 

.-.     (27rx0)  ffdxdy  =  ff(27rx)dxdy 
or 

ff(2Trx)dxdy    =  2TrxQffdxdy. 

But  ff(2irx)dxdy  =  volume  described  in  one  revolution,  and 
2irxQ  =  path  described  by  the  centre  of  gravity  in  one  revolu- 
tion. Hence  follows  the  proposition. 

The  same  propositions  hold  true  for  any  part  of  a  revolution, 
as  well  as  for  an  entire  revolution,  since  we  might  have  multi- 
plied through  by  the  circular  measure  0,  instead  of  by  2-n-. 

It  is  evident  that  the  first  of  these  two  theorems  may  be 
used  to  determine  the  centre  of  gravity  of  a  line,  when  the 
length  of  the  line,  and  the  surface  described  by  revolving  it 
about  the  axis,  are  known  ;  and  so  also  that  the  second  theorem 
may  be  used  to  determine  the  centre  of  gravity  of  a  plane  area 
whenever  the  area  is  known,  and  also  the  volume  described  by 
revolving  it  around  the  axis. 

EXAMPLES. 

i.  Circular  Arc  AC  (Fig.  138). — Length  of  arc  =  s  —  2rO,  sur- 
face of  zone  described  by  revolving  it  about  OY  ==  circumference  of  a 
great  circle  multiplied  by  the  altitude  =  (2irr)  (2rsin0x); 

01)  /.     x00t  =  rsintf, 

sin0r 


236  APPLIED   MECHANICS. 

2.  Semicircular  Arc  (Fig.  139).  —  Length  of  arc  =  irr,  surface  of 
sphere  described  =  4-n-r2  ; 


x    =  — 


3.  Trapezoid  (Fig.  147).  —  Let  AD  =  b,  EC  =  b;  let  it  revolve 
around  AD  :  it  generates  two  cones  and  a  cylinder. 


..       AD  +  BC    „ 
Area  of  trapezoid  =  BG, 

Volume  =  — -(AG  +  HD)  +  -n-(GB)2 .  BC 


L 


HD+ 


FlG-  J47-  =  ^ i-  (AD  +  BC  +  ^C) 


3'IH 


3 
6^^ 


4.    Circular  Sector  AGO    (Fig.  146).  —  Area   of  sector  =   r 
volume   described  =  -^(surface   of  zone)   =   Jr(«nr)  (ar  sin  0,) 
sin  ^!  ; 

sin  Ot 


§165.  Centre  of  Gravity  of  Solid  Bodies.  —  The  general 
formulae  furnish,  in  most  cases,  a  very  complicated  solution,  anc1 
hence  we  generally  have  recourse  to  some  simpler  method.  A 
few  examples  will  be  given  in  this  and  the  next  section. 


CENTRE   OF  GRAVITY  OF  SYMMETRICAL   BODIES.      237 


Tetrahedron  ABCD  (Fig.  148).— The  plane  ABE,  containing  the 
edge  AB  and  the  middle  point  E  of  the  edge  CD,  bisects  all  lines 
drawn  parallel  to  CD,  and  terminating  in  the  faces  A 

ABD  and  ABC :  hence  a  similar  reasoning  to  that 
used  in  the  case  of  the  triangle  will  show  that  the  cen- 
tre of  gravity  of  the  pyramid  must  be  in  the  plane 
ABE  ;  in  the  same  way  it  may  be  shown  that  it  must 
lie  in  the  plane  ACF.  Hence  it  must  lie  in  their 
intersection,  or  in  the  line  AG  joining  the  vertex  A 
with  the  centre  of  gravity  (intersection  of  the  medians) 
of  the  opposite  face. 


FIG.  148. 
In  the  same  way  it  can  be  shown  that  the  centre 


of  gravity  of  the  triangular  pyramid  must  lie  in  the  line  drawn  from 
the  vertex  B  to  the  centre  of  gravity  of  the  face  A  CD.  Hence  the 
centre  of  gravity  of  the  tetrahedron  will  be  found  on  the  line  AG  at 
a  distance  from  G  equal  to 


§  1 66.  Centre  of  Gravity  of  Bodies  which  are  Symmet- 
rical with  Respect  to  an  Axis.  —  Such  solids  may  be  gener- 
z  ated  by  the  motion  of  a  plane  figure,  as  ABCD 

(Fig.  149),  of  variable  dimensions,  and  of  any 
form  whose  centre  G  remains  upon  the  axis 
OX ;  its  plane  being  always  perpendicular  to 
OX,  and  its  variable  area  X  being  a  function 
of  x,  its  distance  from  the  origin. 

Here  the  centre  of  gravity  will  evidently 
FIG.  i49.  lie  on  the  axis  OX,  and  the  elementary  vol- 

ume will  be  the  volume  of  a  thin  plate  whose  area  is  X  and 
thickness  A;r;  hence  the  elementary  volume  will  be 


Take  moments  about  O  Y,  and  we  shall  have 

x0fXdx  =  fXxdx     and     Volume  =  fXdx, 
fXxdx 


or 


fXdx 


V  =  fXdx. 


238 


APPLIED   MECHANICS. 


EXAMPLES. 

x2       y2       z2 

i.  Ellipsoid  —  -f  73  +  -j  =  i  (Fig.  150).  —  Find  centre  of  gravity 
a        D        c 

z  of  the  half  to  the  right  of  the  x  plane.    Let  OK 

=  x.     Now  if,  in  the  equation  of  the  ellipsoid, 

x  X2          Z2 

we  make  y  =  o,  we  have  —  +  —  =  i ; 

c 
~~  a 


where  z  = 


Make  z  =  o  in  the  equation  of  the  ellipsoid,  and  —  +  —  =  i ; 


where  y  = 


/ . 

/.     EK  =  -Va2  —  x2,  KG  =  -\a2  -  x2, 

a  a 


are  the  semi-axes  of  the  variable  ellipse  EGFH,  which,  by  moving  along 
OX,  generates  the  ellipsoid.     Hence 


hence 


Area  EGFH  =  Tr(EK .  GK)  =  —  (a2  -  x2)  =  X; 

irbc 
Elementary  volume  =  — ^  (a2  —  x2)Ax 

-rrbc  C"  ,  .  :';'  .          j  a2.*2  _  ^)a 

(2       "4)0 


azx  — 


#0     = 


3  'o 


irbc  Ta  ,  .    ,  (  X* ) 

I     (a2  -  y*)dx  \a2x--\ 

a2  Jo  (  3  ' 

V  =  —  /     (a2  -  x2)dx     =  \irabc. 
a2  Jo 

2.  Hemisphere.  —  Make  a  =  b  =  t,  and  x0  =  f  <z,  V 


CENTRE   OF  GRAVITY  OF  SYMMETRICAL   BODIES.      2$$ 

If  the  section  X  were  oblique  to  OX,  making  an  angle  6 
with  it,  the  elementary  volume  would  not  be  Xdx,  but  Xdx  sin  0, 
and  we  should  have 

smOfXxdx       fXxdx          ,      T.        •    nrvj 
x0  =  -  —irv,     =     fv,       and     V  =  sin  Of  Xdx. 
sin  Of  Xdx         fXdx 

3.  Oblique  Cone  (Fig.  151).  —  Let  OA  —  h;  let  area  of  base  be 
v4,  and  let  the  angle  made  by  OX  with  the  base  be  0  ; 

X      x2  A 


FIG.  151. 


A       rh 

V  =  sin  Of  Xdx  -  j-  sin  0  I    x2dx  =  \Ah  sin  6. 

9*9 

4.  Truncated  Cone   (Fig.  151).  —  Let   height   of  entire  cone  be 
h  —  OA  ;  let  height  of  portion  cut  off  be  hl ; 

A  rh   ,  h*  -  h* 

_  /     x*dx       - 

4  ~h*>  —  h£ 


A 

TtJk, 3 


24O  APPLIED   MECHANICS. 


CHAPTER  VI. 
STRENGTH  OF  MATERIALS. 

§  167.  Stress,  Strain,  and  Modulus  of  Elasticity.  —  When 
a  body  is  subjected  to  the  action  of  external  forces,  if  we 
imagine  a  plane  section  dividing  the  body  into  two  parts,  the 
force  with  which  one  part  of  the  body  acts  upon  the  other 
at  this  plane  is  called  the  stress  on  the  plane  ;  and,  in  order 
to  know  it  completely,  we  must  know  its  distribution  and  its 
direction  at  each  point  of  the  plane.  If  we  consider  a  small 
area  lying  in  this  plane,  including  the  point  O,  and  represent 
the  stress  on  this  area  by  /,  whereas  the  area  itself  is  repre- 
sented by  a,  then  will  the  limit  of  <-  as  a  approaches  zero  be  the 

a 

intensity  of  the  stress  on  the  plane  under  consideration  at  the 
point  0. 

When  a  body  is  subjected  to  the  action  of  external  forces, 
and,  in  consequence  of  this,  undergoes  a  change  of  form,  it 
will  be  found  that  lines  drawn  within  the  body  are  changed,  by 
the  action  of  these  external  forces,  in  length,  in  direction,  or 
in  both;  and  the  entire  change  of  form  of  the  body  may  be 
correctly  described  by  describing  a  sufficient  number  of  these 
changes. 

If  we  join  two  points,  A  and  B,  of  a  body  before  the 
external  forces  are  applied,  and  find,  that,  after  the  application 
of  the  external  forces,  the  line  joining  the  same  two  points  of 
the  body  has  undergone  a  change  of  length  &(AJ3),  then  is  the 


STRESS,   STRAIN,   AND   MODULUS   OF  ELASTICITY.      241 


limit  of   the  ratio  —  -  -  -  as  AB  approaches   zero   called   the 


strain  of  the  body  at  the  point  A  in  the  direction  AB. 

If  AB  +  k(AB)  >  AB,  the  strain  is  one  of  tension. 

If  AB  4-  k(AB)  <  AB,  the  strain  is  one  of  compression. 

Suppose  a  straight  rod  of  unifo?m  section  A  to  be  subjected 
to  a  pull  P  in  the  direction  of  its  length,  and  that  this  pull  is 
uniformly  distributed  over  the  cross-section  :  then  will  the  in- 
tensity of  the  stress  on  the  cross-section  be 


If  P  be  measured  in  pounds,  and  A  in  square  inches,  then  will 
/  be  measured  in  pounds  per  square  inch. 

If  the  length  of  the  rod  before  the  load  is  applied  be  /, 
and  its  length  after  the  load  is  applied  be  /  +  ^,  then  is  e  the 
elongation  of  the  rod;  and  if  this  elongation  is  uniform  through- 
out the  length  of  the  rod,  then  is  -  the  elongation  of  the  rod 

per  unit  of  length,  or  the  strain. 

Hence,  if  a  represent  the  strain  due  to  the  stress  /  per 
unit  of  area,  we  shall  have 

_  e 

The  Modulus  of  Elasticity  is  the  quotient  obtained  by 
dividing  the  stress  per  unit  of  area  by  the  strain,  or 

E  =  p-, 


and  this  is  expressed  in  units  of  weight  per  unit  of  area,  as  in 
pounds  per  square  inch. 


242  APPLIED   MECHANICS. 

The  modulus  of  elasticity  is  sometimes  defined  as  the 
weight  that  would  be  required  to  stretch  a  rod  one  square  inch 
in  section  to  double  its  length,  if  Hooke's  law,  "The  stress  is 
proportional  to  the  strain,"  held  up  to  that  point,  and  the  rod 
did  not  break.  Of  course  the  modulus  of  elasticity  is  a  con- 
stant for  any  given  material  just  as  far  as,  and  no  farther  than, 
Hooke's  law  holds.  f 

EXAMPLES. 

1.  A  wrought-iron  rod  10  feet  long  and  i  inch  in  diameter  is  loaded 
in  the  direction  of  its  length  with  8000  Ibs. ;  find  (i)  the  intensity  of 
the  stress,  (2)  the  elongation  of  the  rod;  assuming  the  modulus  of  the 
iron  to  be  28000000  Ibs.  per  square  inch. 

2.  What  would  be  the  elongation   of  a    similar  rod   of  cast-iron 
under  the  same  load,  assuming  the  modulus  of  elasticity  of  cast-iron  to 
be  1 7000000  Ibs.  per  square  inch  ? 

3.  Given  a  steel  bar,  area  of  section  being  4  square  inches,  the 
length  of  a  certain  portion  under  a  load  of  25000  Ibs.  being  10  feet, 
and  its  length  under  a  load  of  100000  Ibs.  being  10'  0^.075  ;  find  the 
modulus  of  elasticity  of  the  material. 

4.  What  load  will  be  required  to  stretch  the  rod  in  the  first  example 
Y1^  inch  ? 

§  1 68.  Resistance  to  Stretching  and  Tearing.  —  The  most- 
used  criterion  of  safety  against  injury  for  a  loaded  piece  is, 
that  the  greatest  intensity  of  the  stress  to  which  any  part  of  it 
is  subjected  shall  nowhere  exceed  a  certain  fixed  amount,  called 
the  working-strength  of  the  material  ;  this  working-strength 
being  a  certain  fraction  of  the  breaking-strength  determined 
by  practical  considerations. 

The  more  correct  but  less  used  criterion  is,  that  the  great- 
est strain  in  any  part  of  the  structure  shall  nowhere  exceed 
the  working-strain ;  the  greatest  allowable  amount  of  strain 
being  a  fixed  quantity  determined  by  practical  considerations. 


RESISTANCE    TO  STRETCHING   AND    TEARING.          243 

This  is  equivalent  to  limiting  the  allowable  elongation  or 
compression  to  a  certain  fraction  of  its  length,  or  the  deflection 
of  a  beam  to  a  certain  fraction  of  the  span. 

If  the  stress  on  a  plane  surface  be  uniformly  distributed, 
its  resultant  will  evidently  act  at  the  centre  of  gravity  of  the 
surface,  as  has  been  already  shown  in  §  42  to  be  the  case  with 
any  uniformly  distributed  force. 

If  a  straight  rod  of  uniform  section  and  material  be  sub- 
jected to  a  pull  in  the  direction  of  its  length,  and  if  the  result- 
ant of  the  pull  acts  along  a  line  passing  through  the  centres 
of  gravity  of  the  sections  of  the  rod,  it  is  assumed  in  practice 
that  the  stress  is  uniformly  distributed  throughout  the  rod,  and 
hence  that  for  any  section  we  shall  obtain  the  stress  per  square 
inch  by  dividing  the  total  pull  by  the  number  of  square  inches 
in  the  section. 

If,  on  the  other  hand,  the  resultant  of  the  pull  does  not  act 
through  the  centres  of  gravity  of  the  sections,  the  pull  is  not 
uniformly  distributed ;  and  while 


will  express  the  mean  stress  per  square  inch,  the  actual  inten- 
sity of  the  stress  will  vary  at  different  points  of  the  section, 

p 
being  greater  than  —  at  some  points  and  less  at  others.     How 

yi 

to  determine  its  greatest  intensity  in  such  cases  will  be  shown 
later. 

With  good  workmanship  and  well-fitting  joints,  the  first 
case,  or  that  of  a  uniformly  distributed  stress,  can  be  practi- 
cally realized ;  but  with  ill-fitting  joints  or  poor  workmanship, 
or  with  a  material  that  is  not  homogeneous,  the  resultant  of 
the  pull  is  liable  to  be  thrown  to  one  side  of  the  line  passing 
through  the  centres  of  gravity  of  the  sections,  and  thus  there 


244  APPLIED   MECHANICS. 

is  set  up  a  bending-action  in  addition  to  the  direct  tension,  and 
therefore  an  unevenly  distributed  stress. 

It  is  of  the  greatest  importance  in  practice  to  take  cogni- 
zance of  any  such  irregularities,  and  determine  the  greatest 
intensity  of  the  stress  to  which  the  piece  is  subjected :  though 
it  is  too  often  taken  account  of  merely  by  means  of  a  factor  of 
safety  ;  in  other,  words,  by  guess. 

Leaving,  then,  this  latter  case  until  we  have  studied  the 
stresses  due  to  bending,  we  will  confine  ourselves  to  the  case 
of  the  uniformly  distributed  stress. 

If  the  total  pull  on  the  rod  in  the  direction  of  its  length 
be  P,  and  the  area  of  its  cross-section  A,  we  shall  have,  for  the 
intensity  of  the  pull, 


On  the  other  hand,  if  the  working-strength  of  the  material 
per  unit  of  area  be  f,  we  shall  have,  for  the  greatest  admissible 
load  to  be  applied, 

If  f  be  the  working-strength  of  the  material  per  square 
inch,  and  E  the  modulus  of  elasticity,  then  is  the  greatest 
admissible  strain  equal  to 

a  =  £- 

Thus,  assuming  12000  Ibs.  per  square  inch  as  the  working 
tensile  strength  of  wrought-iron,  and  28000000  Ibs.  per  square 
inch  as  its  modulus  of  elasticity,  its  working-strain  would  be 

12000  3 

a  = 


28OOOOOO          7000 

Hence  the  greatest  safe  elongation   of   the   bar  would   be 
of  its  length.     Hence  a  rod   10  feet  long  could  safely  be 
stretched  T|¥  of  a  foot  =  0.05 14". 


VALUES  OF  BREAKING  AND    WORKING   STRENGTH.    245 

§  169.  Approximate  Values  of  Breaking  Strength,  and 
of  Modulus  of  Elasticity. — In  a  later  part  of  this  book  the 
attempt  will  be  made  to  give  an  account  of  the  experiments 
that  have  been  made  to  determine  the  strength  and  elas- 
ticity of  the  materials  ordinarily  used  in  construction,  in  such 
a  way  as  to  enable  the  student  to  decide  for  himself,  in  any 
special  case,  upon  the  proper  values  of  the  constants  that  he 
ought  to  use. 

For  the  present,  however,  the  following  will  be  given  as  a 
rough  approximation  to  some  of  these  quantities,  which  we  may 
make  use  of  in  our  work  until  we  reach  the  above-mentioned 
account. 

(a)  Cast-Iron. 

Breaking  tensile  strength  per  square  inch,  of  common  quali- 
ties, 14000  to  20000  Ibs. ;  of  gun  iron,  30000  to  33000  Ibs. 

Modulus  of  elasticity  for  tension  and  for  compression,  about 
17000000  Ibs.  per  square  inch. 

(b)  Wrought- Iron. 

Breaking  tensile  strength  per  square  inch,  from  40000  to 
60000  Ibs. 

Modulus  of  elasticity  for  tension  and  for  compression,  about 
28000000. 

(c)  Mild  Steel. 

Breaking  tensile  strength  per  square  inch,  55000  to  70000 
Ibs. 

Modulus  of  elasticity  for  tension  and  for  compression,  from 
28000000  to  30000000  Ibs.  per  square  inch. 

(d)  Wood. 

Breaking  compressive  strength  per  square  inch  :  — 

Oak,  green     .......  3000  Ibs. 

Oak,  dry 3000  to  6000  Ibs. 

Yellow  pine,  green 3000  to  4000  Ibs. 

Yellow  pine,  dry 4000  to  7000  Ibs. 


246 


APPLIED  MECHANICS. 


Modulus  of  elasticity  for  compression  (average  values) :  — 

Oak .    1300000  Ibs.  per  square  inch. 

Yellow  pine 1600000  Ibs.  per  square  inch. 

§  170.  Sudden  Application  of  the  Load. —  If  a  wrought- 
iron  rod  10  feet  long  and  I  square  inch  in  section  be  loaded 
with  12000  pounds  in  the  direction  of  its  length,  and  if  the 
modulus  of  elasticity  of  the  iron  be  28000000,  it  will  stretch 
0.05 14"  provided  the  load  be  gradually  applied :  thus,  the  rod 
begins  to  stretch  as  soon  as  a  small  load  is  applied  ;  and,  as  the 
load  gradually  increases,  the  stretch  increases,  until  it  reaches 
0.05 1 4". 

If,  on  the  other  hand,  the  load  of  12000  Ibs.  be  suddenly 
applied  (i.e.,  put  on  all  at  once)  without  being  allowed  to  fall 
through  any  height  beforehand,  it  would  cause  a  greater  stretch 
at  first,  the  rod  undergoing  a  series  of  oscillations,  finally 
settling  down  to  an  elongation  of  0.05 14". 

To  ascertain  what  suddenly  applied  load  will  produce  at 
most  the  elongation  0.0514",  observe,  that,  in  the  case  of  the 
gradually  applied  load,  we  have  a  load  gradually  increasing  from 

o  to  12000  Ibs. 

Its  mean  value  is,  therefore,  \(  12000)  =  6000  Ibs.  ;  and  this 
force  descends  through  a  distance  of 

0.0514". 

Hence  the  amount  of  mechanical  work  done  on  the  rod  by  the 
gradually  applied  load  in  producing  this  elongation  is 

(6000)  (0.0514)  =  308.4  inch-lbs. 

Hence,  if  we  are  to  perform  upon  the  rod  308.4  inch-lbs.  of 
work  with  a  constant  force,  and  if  the  stretch  is  to  be  0.05 14", 
the  magnitude  of  the  force  must  be 

3°8-4,,  =  6000  Ibs. 

O.O5I4 


RESILIENCE    OF  A    TENSION-BAR.  247 

Hence  a  suddenly  applied  load  will  produce  double  the  strain 
that  would  be  produced  by  the  same  load  gradually  applied  ; 
and,  moreover,  a  suddenly  applied  load  should  be  only  half  as 
great  as  one  gradually  applied  if  it  is  to  produce  the  same 
strain. 

§  171.  Resilience  of  a  Tension-Bar.  —  The  resilience  of  a 
tension-rod  is  the  mechanical  work  done  in  stretching  it  to  the 
same  amount  that  it  would  stretch  under  the  greatest  allowable 
gradually  applied  load,  and  is  found  by  multiplying  the  greatest 
allowable  load  by  half  the  corresponding  elongation. 

Thus,  suppose  a  load  of  100  Ibs.  to  be  dropped  upon  the 
rod  described  above  in  such  a  way  as  to  cause  an  elongation  not 
greater  than  0.05 14",  it  would  be  necessary  to  drop  it  from  a 
height  not  greater  than  3.08". 

EXAMPLES. 

1 .  A  wrought-iron  rod  is  1 2  feet  long  and  T  inch  in  diameter,  and 
is  loaded  in  the  direction  of  its  length ;   the  working-strength   of  the 
iron  being  12000  Ibs.  per  square  inch,  and  the  modulus  of  elasticity 
28000000  Ibs.  per  square  inch. 

Find  the  working-strain. 
Find  the  working-load. 
Find  the  working-elongation. 
Find  the  working-resilience. 

From  what  height  can  a  5o-pound  weight  be  dropped  so  as  to  produce 
tension,  without  stretching  it  more  than  the  working- elongation? 

2.  Do  the  same  for  a  cast-iron  rod,  where  the  working-strength  is 
5000  pounds  per  square  inch,  and  the  modulus  of  elasticity  17000000; 
the  dimensions  of  the  rod  being  the  same. 

§  172.  Results  of  Wohler's  Experiments  on  Tensile 
Strength.  —  According  to  the  experiments  of  Wohler,  of  which 
an  account  will  be  given  later,  the  breaking-strength  of  a  piece 


248  APPLIED   MECHANICS. 

depends,  not  only  on  whether  the  load  is  gradually  or  suddenly 
applied,  but  also  on  the  extreme  variations  of  load  that  the 
piece  is  called  upon  to  undergo,  and  the  number  of  changes  to 
which  it  is  to  be  submitted  during  its  life. 

For  a  piece  which  is  always  in  tension,  he  determines  the 
following  two  constants  ;  viz.,  t,  the  carrying-strength  per  square 
inch,  or  the  greatest  quiescent  stress  that  the  piece  will  bear, 
and  u,  the  primitive  safe  strength,  or  the  greatest  stress  per 
square  inch  of  which  the  piece  will  bear  an  indefinite  number 
of  repetitions,  the  stress  being  entirely  removed  in  the  inter- 
vals. 

This  primitive  safe  strength,  u,  is  used  as  the  breaking- 
strength  when  the  stress  varies  from  o  to  u  every  time.  Then, 
by  means  of  Launhardt's  formula,  we  are  able  to  determine  the 
ultimate  strength  per  square  inch  for  any  different  limits  of 
stress,  as  for  a  piece  that  is  to  be  alternately  subjected  to  80000 
and  6000  pounds. 

Thus,  for  Phoenix  Company's  axle  iron,  Wohler  finds 

t  =  3290  kil.  per  sq.  cent.  =  46800  Ibs.  per  sq.  in., 
u  =  2100  kil.  per  sq.  cent.  =  30000  Ibs.  per  sq.  in. 

Launhardt's  formula  for  the  ultimate  strength  per  unit  of  area 


s 


!t  —  u     least  stress    | 
u      greatest  stress)* 


Hence,  with  these  values  of  t  and  «,  we  should  have,  for  the 
ultimate  strength  per  square  inch, 

(          i    least  stress    )    . 

#=     2ioo-,'i+-  ->kil.  per  sq.  cent., 

{          2 greatest  stress) 

or 

Si    least  stress    ) 
i  + -  >  Ibs.  per  sq.  in. 

2  greatest  stress  ( 


WOHLER'S  EXPERIMENTS  ON  TENSILE  STRENGTH.      249 

Thus,  if  least  stress  =  6000,  and  greatest  =  80000,  we  should 

have 

a  =  30000^1  +  i-  &j  =  30000^1  +  tfU 


if  least  stress  =  60000,  and  greatest  =  80000, 

a  =  30000 \  i  -f-  -J  .  % \  =  30000^1  -f-  H§  =  41250; 

if  least  stress  =  greatest  stress  =  80000, 

a  =  30000  \  i  +  %\  =  45000  =  carrying-strength. 

Hence,  instead  of  using,  as  breaking-strength  per  square  inch 
in  all  cases,  45000,  we  should  use  a  set  of  values  varying  from 
45000  down  to  30000,  according  to  the  variation  of  stress  which 
the  piece  is  to  undergo. 

For  working-strength,  Weyrauch  divides  this   by  3  :  thus 
obtaining,  for  working-strength  per  square  inch, 

(      ,    i     least  stress    )  „ 

b  =  10000  \  i  -\ [  Ibs.  per  sq.  in. ; 

(         2  greatest  stress  j 

for  Krupp's  cast-steel,  although 

t  =  7340  kil.  per  sq.  cent.  =  104400  Ibs.  per  sq.  in., 
u  =  3300  kil.  per  sq.  cent.  =    46900  Ibs.  per  sq.  in., 

Weyrauch  recommends 

(          q     least  stress    ) 

a  =     3300  \  i  -f-  — -  >  kil.  per  sq.  cent., 

00      (          ii  greatest  stress) 

or 

(  q     least  stress    )  , 

a  =  46000  <  i  4-  — >lbs.  per  sq.  m., 

(          ii  greatest  stress  ( 

(  9     least  stress    ) 

/.     b  =  15633  <  i  +  —  -  ->lbs.  per  sq.  in. 

T|          "greatest  stress) 

EXAMPLES. 

Find  the  breaking-strength  per  square  inch  for  a  wrought-iron  tension 
rod. 

1.  Extreme  loads  are     75000  and      6000  Ibs. 

2.  Extreme  loads  are  120000  and  100000  Ibs. 

3.  Extreme  loads  are  300000  and     TOOOO  Ibs. 
Find  the  safe  section  for  the  rod  in  each  case-, 


0*  THR 

'UNIVERSITY; 


25O  APPLIED   MECHANICS. 

§173.  Suspension- Rod  of  Uniform  Strength.  —  In  the 
case  of  a  long  suspension-rod,  the  weight  of  the  rod  itself  some- 
times becomes  an  important  item.  The  upper  section  must,  of 
course,  be  large  enough  to  bear  the  weight  that  is  hung  from 
the  rod  plus  the  weight  of  the  rod  itself;  but  it  is  sometimes 
desirable  to  diminish  the  sections  as  they  descend.  This  is  often 
accomplished  in  mines  by  making  the  rod  in  sections,  each  section 
being  calculated  to  bear  the  weight  below  it  plus  its  own  weight. 
Were  the  sections  gradually  diminished,  so  that  each  section 
would  be  just  large  enough  to  support  the  weight  below  it,  we 
should,  of  course,  have  a  curvilinear  form ;  and  the  equation  of 
this  curve  could  be  found  as  follows,  or,  rather,  the  area  of  any 
section  at  a  distance  from  the  bottom  of  the  rod. 
Let  W  =  weight  hung  at  O  (Fig.  152), 

Let  w  =  weight  per  unit  of  volume  of 

the  rod, 
(\-\-f  Let  x  =  distance  AO, 

Let  5  =  area  of  section  A, 

Let  x  -f-  dx  =  distance  BO, 
Let  5  +  dS  =  area  of  section  at  B, 
Let  f  —  working-strength  of  the  mate- 

rial per  square  inch. 

i°.  The  section  at  O  must  be  just  large  enough 
to  sustain  the  load  W ; 


o 


W 


FIG.  152. 


2°.  The  area  in  dS  must  be  just  enough  to  sustain   the 
weight  of  the  portion  of  the  rod  between  A  and  B. 
The  weight  of  this  portion  is  wSdx  ; 

wSdx 


'*= 


dS       w  w 

-=  =  -jdx  /.    log*  £  =  -fX  -h  a  constant. 

M       /  / 


CYLINDERS  SUBJECTED    TO   INTERNAL   PRESSURE.     251 


W 

When  x  —  o,  5  —  —=  ; 

W  IW\       w 

-y  =  the  constant  /.    log^o  —  log*  I  — 7- )  =  -^ 


This  gives  us  the  means  of  determining  the  area  at  any  dis- 
tance x  from  (9. 


1.  A  wrought-iron  tension-rod  200  feet  long  is  to  sustain  a  load  of 
2000  Ibs.  with  a  factor  of  safety  of  4,  and  is  to  be  made  in  4  sections, 
each  50  feet  long  ;  find  the  diameter  of  each  section,  the  weight  of  the 
wrought  iron  being  480  Ibs.  per  cubic  foot. 

2.  Find  the  diameter  needed  if  the  rod  were   made   of  uniform 
section,  also  the  weight  of  the  extra  iron  necessary  to  use  in  this  case. 

3.  Find  the  equation  of  the  longitudinal  section  of  the  rod,  assum- 
ing a  square  cross-section,  if  it  were  one  of  uniform  strength,  instead  of 
being  made  in  4  sections. 

§  174.  Thin  Hollow  Cylinders  subjected  to  an  Internal 
Normal  Pressure.  —  Let/  denote  the  uniform  intensity  of  the 
pressure  exerted  by  a  fluid  which  is  confined  within  a  hollow 
cylinder  of  radius  r  and  of  thickness  t  (Fig,  153), 
the  thickness  being  small  compared  with  the  radius. 
Let  us  consider  a  unit  of  length  of  the  cylinder,  and  c( 
let  us  also  consider  the  forces  acting  on  the  upper 
half-ring  CED. 

The  total  upward  force  acting  on  this  half-ring,  in  conse- 
quence of  the  internal  normal  pressure,  will  be  the  same  as 
that  acting  on  a  section  of  the  cylinder  made  by  a  plane  pass- 
ing through  its  axis,  and  the  diameter  CD.  The  area  of  this 


252  APPLIED   MECHANICS. 

section  will  be  2r  X  I  =  2r :  hence  the  total  upward  force  will  be 
2/"  X/  —  2pr;  and  the  tendency  of  this  upward  force  is  to  cause 
the  cylinder  to  give  way  at  A  and  B,  the  upper  part  separating 
from  the  lower. 

This  tendency  is  resisted  by  the  tension  in  the  metal  at  the 
sections  AC  and  BD ;  hence  at  each  of  these  sections,  there  has 
to  be  resisted  a  tensile  stress  equal  to  \(zpr)  =  pr.  This  stress 
is  really  not  distributed  uniformly  throughout  the  cross-section 
of  the  metal ;  but,  inasmuch  as  the  metal  is  thin,  no  serious 
error  will  be  made  if  it  be  accounted  as  distributed  uniformly. 
The  area  of  each  section,  however,  is  /  X  i  =£/  therefore,  if 
T  denote  the  intensity  of  the  tension  in  the  metal  in  a  tangential 
direction  (i.e.,  the  intensity  of  the  hoop  tension),  we  shall  have 


Hence,  to  insure  safety,   T  must  not  be  greater  than  /,  the 
working-strength  of  the  material  for  tension  ;  hence,  putting 

/=-> 
we  shall  have 


as  the  proper  thickness,  when/  =  normal  pressure  per  square 
inch,  and  radius  =  r. 

The  above  are  the  formulae  in  common  use  for  the  deter- 
mination of  the  thickness  of  the  shell  of  a  steam-boiler ;  for  in 
that  case  the  steam-pressure  is  so  great  that  the  tension 
induced  by  any  shocks  that  are  likely  to  occur,  or  by  the  weight 
of  the  boiler,  is  very  small  in  comparison  with  that  induced 
by  the  steam-pressure.  On  the  other  hand,  in  the  case  of  an 
ordinary  water-pipe,  the  reverse  is  the  case. 


RESISTANCE    TO  DIRECT  COMPRESSION. 


To  provide  for  this  case,  Weisbach  directs  us  to  add  to  the 
thickness  we  should  obtain  by  the  above  formulae,  a  constant 
minimum  thickness. 

The  following  are  his  formulae,  d  being  the  diameter  in 
inches,  p  the  internal  normal  pressure  in  atmospheres,  and  t 
the  thickness  in  inches.  For  tubes  made  of 

Sheet-iron  .........  /=  0.00086  pd  +  0.12 

Cast-iron     .........  /  =  0.00238^  -j-  0.34 

Copper  ..........  /  =  0.00148/tf7  -f  0.16 

Lead      ..........  /=  0.00507/^/4-  0.21 

Zinc  ........     „../==  o.  00242  pd  -f-  0.16 

Wood     ..........  t  =  0.03230^  -f  1.07 

Natural  stone  ........  /  =  0.03690  pd  -f-  1.18 

Artificial  stone      .......  /  =  0.05380^  -f-  1.58 

§  175.  Resistance  to  Direct  Compression.  —  When  a  piece 
is  subjected  to  compression,  the  distribution  of  the  compressive 
stress  on  any  cross-section  depends,  first,  upon  whether  the 
resultant  of  the  pressure  acts  along  the  line  containing  the  cen- 
tres of  gravity  of  the  sections,  and,  secondly,  upon  the  dimen- 
sions of  the  piece  ;  thus  determining  whether  it  will  bend  or 
not. 

In  the  case  of  an  eccentric  load,  or  of  a  piece  of  such  length 
that  it  yields  by  bending,  the  stress  is  not  uniformly  distributed  ; 
and,  in  order  to  proportion  the  piece,  we  must  determine  the 
greatest  intensity  of  the  stress  upon  it,  and  so  proportion  it 
that  this  shall  be  kept  within  the  working-strength  of  the  ma- 
terial for  compression. 

Either  of  these  cases  is  not  a  case  of  direct  compres- 
sion. 

In  the  case  of  direct  compression  (i.e.,  where  the  stress  over 
each  section  is  uniformly  distributed),  the  intensity  of  the  stress 
is  found  by  dividing  the  total  compression  by  the  area  of  the 


254  APPLIED   MECHANICS. 

section  ;  so  that,  if  P  be  the  total  compression,  and  A  the  area 
of  the  section,  and/  the  intensity  of  the  compressive  stress, 


On  the  other  hand,  if  /  is  the  compressive  working-strength  oi 
the  material  per  square  inch,  and  A  the  area  of  the  section  in 
square  inches,  then  the  greatest  allowable  load  on  the  piece 
subjected  to  compression  is 


The  same  remarks  as  were  made  in  regard  to  a  suddenly 
applied  load  and  resilience,  in  the  case  of  direct  tension,  apply 
in  the  case  of  direct  compression. 

§  176.  Results  of  Wohler's  Experiments  on  Compressive 
Strength.  —  Wohler  also  made  experiments  in  regard  to  pieces 
subjected  to  alternate  tension  and  compression,  taking,  in  the 
experiments  themselves,  the  case  where  the  metal  is  subjected 
to  alternate  tensions  and  compressions  of  equal  amount. 

The  greatest  stress  of  which  the  piece  would  bear  an  indefi- 
nite number  of  changes  under  these  conditions,  is  called  the 
vibration  safe  strength,  and  is  denoted  by  s. 

Weyrauch  deduces  a  formula  similar  to  that  of  Launhardt 
for  the  greatest  allowable  stress  per  unit  of  area  on  the  piece 
when  it  is  subjected  to  alternate  tensions  and  compressions  of 
different  amounts. 

Thus,  for  Phoenix  Company's  axle  iron,  Wohler  deduces 

/  =  3290  kil.  per  sq.  cent.  =  46800  Ibs.  per  sq.  in., 
u  =  2100  kil.  per  sq.  cent.  =  30000  Ibs.  per  sq.  in., 
s  =  1170  kil.  per  sq.  cent.  =  16600  Ibs.  per  sq.  in. 


EXPERIMENTS  ON  COMPRESSIVE   STRENGTH.  255 

Weyrauch's  formula  for  the  ultimate  strength  per  unit  of 
area  is 

1u  —  s     least  maximum  stress    ) 
i : > : 
u      greatest  maximum  stress  J 

and,  with  these  values  of  «  and  s,  it  gives 

(          i     least  maximum  stress 

a  =     2ioo<  i  —  -  — : — 

(          2  greatest  maximum  stress 

or 

least  maximum  stress 


!i     least  maximum  stress    ) 
i  —  -  —  — : —         —  >  Ibs.  per  sq.  in. 

2  greatest  maximum  stress ) 

With  a  factor  of  safety  of  3,  we  should  have,  for  the  greatest 
admissible  stress  per  square  inch, 

(          i     least  maximum  stress    ) 

b  =  ioooo<  i  — : >lbs. 

(          2  greatest  maximum  stress  ( 

For  Krupp's  cast-steel, 

/  =  7340  kil.  per  sq.  cent.  =  104400  Ibs.  per  sq.  in., 

u  —  33°°  kil.  per  sq.  cent.  =    46900  Ibs.  per  sq.  in.  approximately, 

s   =  2050  kil.  per  sq.  cent.  =     29150  Ibs.  per  sq.  in.  approximately. 

We  have,  therefore,  for  the  breaking-strength  per  unit  of 
area,  according  to  Weyrauch's  formula, 

!c      least  maximum  stress    ) 
1  ~  H  greatest  maximum  stress    kiL  p£r  "*  C6nt" 


or 


(  c      least  maximum  stress    ) 

a  =  46900  <  i  —  f-  -  —  :  —  -  J.  Ibs.  per  sq.  in.  ; 


greatest  maximum  stress 


256  APPLIED   MECHANICS. 

and,  using  a  factor  of  safety  of  3,  we  have,  for  the  greatest  admis- 
sible stress  per  square  inch, 

(  c      least  maximum  stress    ) 

*  =  '5630 1 1  -  n  greatest  maximum  stress  I"*5'  P«r  sq.  in. 

The  principles  respecting  an  eccentric  compressive  load,  and 
those  respecting  the  giving-way  of  long  columns  so  far  as  they 
are  known,  can  only  be  treated  after  we  have  studied  the  resist- 
ance of  beams  to  bending;  hence  this  subject  will  be  deferred 
until  that  time. 

EXAMPLES. 

Find  the  proper  working  and  breaking  strength  per  square  inch  to 
be  used  for  a  wrought- iron  rod,  the  extreme  stresses  being  — 

1.  80000  Ibs.  tension  and      6000  Ibs.  compression. 

2.  100000  Ibs.  tension  and  100000  Ibs.  compression. 

3.  70000  Ibs.  tension  and    60000  Ibs.  compression. 
Do  the  same  for  a  steel  rod. 

§  177.  Resistance  to  Shearing.  —  One  of  the  principal  cases 
where  the  resistance  to  shearing  comes  into  practical  use  is 
that  where  the  members  of  a  structure,  which  are  themselves 
subjected  to  direct  tension  or  compression  or  bending,  are  united 
by  such  pieces  as  bolts,  rivets,  pins,  or  keys,  which  are  sub- 
jected to  shearing.  Sometimes  the  shearing  is  combined  with 
tension  or  with  bending  ;  and  whenever  this  is  the  case,  it  is 
necessary  to  take  account  of  this  fact  in  designing  the  pieces. 
It  is  important  that  the  pins,  keys,  etc.,  should  be  equally 
strong  with  the  pieces  they  connect. 

Probably  one  of  the  most  important  modes  of  connection  is 
by  means  of  rivets.  In  order  that  there  may  be  only  a  shearing 
action,  without  any  bending  of  the  rivets,  the  latter  must  fit 
very  tightly.  The  manner  in  which  the  riveting  is  done  will 
necessarily  affect  very  essentially  the  strength  of  the  joints; 


RESISTANCE    TO   SHEARING. 


257 


hence  the  only  way  to  discuss  fully  the  strength  of  riveted 
joints  is  to  take  into  account  the  manner  of  effecting  the  rivet- 
ing, and  hence  the  results  of  experiments.  These  will  be 
spoken  of  later;  but  the  ordinary  theories  by  which  the  strength 
and  proportions  of  riveted  joints  are  determined  will  be  given, 
which  theories  are  necessary  also  in  discussing  the  results  of 
experiments  thereon. 

The  principle  on  which  the  theory  is  based  is  that  of  making 
the  resistance  of  the  joint  to  yielding  equal  in  the  first  three, 
and  also  in  either  the  fourth  or  the  fifth  of  the  ways  in  which  it 
is  possible  for  it  to  yield,  as  enumerated  on  pages  258  and  259. 


A  single-riveted  lap-joint  is  one 
with  a  single  row  of  rivets,  as 
shown  in  Fig.  154. 


A  single-riveted  butt-joint  with 
one  covering  plate  is  shown  in 
Fig.  155- 


FIG.  154. 


FIG.  155. 


A  single-riveted  butt-joint  with 
two  covering  plates  is  shown  in 
Fig.  156. 


_J 

FIG.  156. 


258 


APPLIED   MECHANICS. 


FIG.  157. 


A  double-riveted  lap-joint  with 
the  rivets  staggered  is  shown  in 
Fig-  157;  one  with  chain  riveting, 
in  Fig.  158. 


FIG.  158. 


Taking  the  case  of  the  single-riveted  lap-joint  shown  in  Fig. 
C54,  it  may  yield  in  one  of  five  ways  :  — 


i°.  By  the  crushing  of  the  plate 
in  front  of  the  rivet  (Fig.  159). 


FIG.  159. 


FIG.  160. 


2°.  By  the  shearing  of  the  rivet 
(Fig.  1 60). 


RESISTANCE    TO   SHEARING. 


259 


3°.  By  the  tearing  of  the  plate 
between  the  rivet-holes  (Fig.  161). 
( 

1 

FIG.  161. 


4°.  By  the  rivet  breaking 
through  the  plate  (Fig.  162). 

5°.  By  the  rivet  shearing  out 
the  plate  in  front  of  it. 


Let  us  call 

d  the  diameter  of  a  rivet. 

c  the  pitch  of  the  rivets  ;  i.e.,  FlG- l62- 

their  distance  apart  from  centre  to  centre. 
t  the  thickness  of  the  plate. 
/  the  lap  of  the  plate  ;  i.e.,  the  distance  from  the  outer  edge 

of  a  rivet-hole  to  the  outer  edge  of  the  plate. 
ft  the  ultimate  tensile  strength  of  the  iron. 
fs  the  ultimate  shearing-strength  of  the  rivet-iron. 
/>  the  ultimate  shearing-strength  of  the  plate. 
fc  the  ultimate  crushing-strength  of  the  iron. 
We  shall  then  have  — 

i°.   Resistance  of  plate  in  front  of  rivet  to  crushing  =  fctd. 

2°.  Resistance  of  one  rivet  to  shearing  =  fi-  -Y 

3°.  Resistance  of  plate  between  two  rivet-holes  to  tearing 

«  ftt(c  -  d). 

tl2 

4°.   Resistance  of  plate  to  being  broken  through  =  a — , 

d 

where  a  is  a  constant  depending  on  the  material.     This  may  be 
taken  as  empirical  for  the  present. 

An  average  value  of  this  constant,  as  given  by  Robert  Wil- 
son, is  100000  Ibs.,  where  all  the  dimensions  are  measured  in 
inches. 


26o»  APPLIED   MECHANICS. 

5°.  Resistance  of  plate  in  front  of  the   rivet  to  shearing 


Assuming  .that  we  know  the  thickness  of  the  plate  to  start 
with,  we  obtain,  by  equating  the  first  two  resistances, 


ftd        f  J 

=  7*T  - 

which  determines  the  diameter  of  the  rivet. 
Equating  3°  and  2°,  we  obtain 


which  gives  the  pitch  of  the  rivets  in  terms  of  the  diameter  of 
the  rivet,  and  the  thickness  of  the  plate. 
Equating,  next,  4°  and  i°,  we  have 

tl2  17 

a-;  =  fctd         :.     I  =  d\J-, 

ft  f  a 

which  gives  the  lap  of  the  plate  needed  in  order  that  it  may  not 
break  through. 

By  equating  5°  and  i°,  we  find  the  lap  needed  that  it  may 
not  shear  out  in  front  of  the  rivet. 

A  similar  method  of  reasoning  would  enable  us  to  determine 
the  corresponding  quantities  in  the  cases  of  double-riveted 
joints,  etc. 

The  above  is  the  ordinary  theory  of  riveted  joints.  There 
are  a  number  of  practical  considerations  which  modify  more  or 
less  the  results  of  this  theory,  and  which  can  only  be  deter- 
mined experimentally.  A  fuller  account  of  this  subject  from 
ail  experimental  point  of  view  will  be  given  later. 

§  178.  Intensity  of  Stress.  —  Whenever  the  stress  over  a 
plane  area  is  uniformly  distributed,  we  obtain  its  intensity  at 
each  point  by  dividing  the  total  stress  by  the  area  over  which  it 
acts,  thus  obtaining  the  amount  per  unit  of  area.  When,  how- 
ever, the  stress  is  not  uniformly  distributed,  or  when  its  inten- 


INTENSITY  OF  STRESS. 


26l 


sity  varies  at  different  points,  we  must  adopt  a  somewhat  differ 
ent  definition  of  its  intensity  at  a  given  point.  In  that  case,  ii 
we  assume  a  small  area  containing  that  point,  and  divide  the 
stress  which  acts  on  that  area  by  the  area,  we  shall  have,  in  the 
quotient,  an  approximation  to  the  intensity  required,  which  will 
approach  nearer  and  nearer  to  the  true  value  of  the  intensity  at 
that  point,  the  smaller  the  area  is  taken. 

Hence  the  intensity  of  a  variable  stress  at  a  given  point  is,  — 

The  limit  of  the  ratio  of  the  stress  acting  on  a  small  area 
containing  that  point,  to  the  area,  as  the  latter  grows  smaller  and 
smaller. 

By  dividing  the  total  stress  acting  on  a  certain  area  by  the 
entire  area,  we  obtain  the  mean  intensity  of  the  stress  for  the 
entire  area. 

§  179.    Graphical  Representation  of  Stress.  —  A  conven- 
ient   mode    of   representing   stress 
graphically  is  the  following:  — 

Let  AB  (Fig.  163)  be  the  plane 
surface  upon  which  the  stress  acts ; 
let  the  axes  OX  and  OY  be  taken 
in  this  plane,  the  axis  OZ  being  at 
right  angles  to  the  plane. 

Conceive  a  portion  of  a  cylinder 
whose  elements  are  all  parallel  to 
OZ,  bounded  at  one  end  by  the 
given  plane  surface,  and  at  the 
other  by  a  surface  whose  ordinate  at  any  point  contains  as 
many  units  of  length  as  there  are  units  of  force  in  the  intensity 
of  the  stress  at  that  point  of  the  given  plane  surface  where  the 
ordinate  cuts  it. 

The  volume  of  such  a  figure  will  evidently  be 

V  =  ffzdxdy  =  ffpdxdy, 
where  z  =  /  —  intensity  of  the  stress  at  the  given  point. 


FIG.  163. 


262 


APPLIED   MECHANICS. 


Hence  the  volume  of  the  cylindrical  figure  will  contain  as 
many  units  of  volume  as  the  total  stress  contains  units  of 
force;  or,  in  other  words,  the  total  stress' will  be  correctly  repre- 
sented by  the  volume  of  the  body. 

If  the  stress  on  the  plane 
figure  is  partly  tension  and 
partly  compression,  the  sur- 
face whose  ordinates  repre- 
sent the  intensity  of  the 
stress  will  lie  partly  on  one 
side  of  the  given  plane  sur- 
face and  partly  on  the  other ; 
this  surface  and  the  plane 
surface  on  which  the  stress 
acts,  cutting  each  other  in 
some  line,  straight  or  curved, 
as  shown  in  Fig.  164.  In  that 
case,  the  magnitude  of  the  resultant  stress  P  =  V  —  ffzdxdy 
will  be  equal  to  the  difference  of  the  wedge-shaped  volumes 
shown  in  the  figure. 

It  will  be  observed  that  the  above  method  of  representing 
stress  graphically  represents,  i°,  the  intensity  at  each  point  of 
the  surface  to  which  it  is  applied  ;  and,  2°,  the  total  amount 
of  the  stress  on  the  surface.  It  does  not,  however,  represent 
its  direction,  except  in  the  case  when  the  stress  is  normal  to 
the  surface  on  which  it  acts. 

In  this  latter  case,  however,  this  is  a  complete  representa- 
tion of  the  stress. 

The  two  most  common  cases  of  stress  are,  i°,  uniform  stress, 
and,  2°,  uniformly  varying  stress.  These  two  cases  are  repre- 
sented respectively  in  Figs.  165  and  166;  the  direction  also 
being  correctly  represented  when,  as  is  most  frequently  the 
case,  the  stress  is  normal  to  the  surface  of  action.  In  Fig. 
165,  AB  is  supposed'  to  be  the  surface  on  which  the  stress 


FIG.  164. 


GRAPHICAL   REPRESENTATION  OF  STRESS. 


263 


acts ;  the  stress  is  supposed  to  be  uniform,  and  normal  to  the 
surface  on  which  it  acts ;  the  bound- 
ing surface  in  this  case  becomes  a 
plane  parallel  to  AB ;  the  intensity 
of  the  stress  at  any  point,  as  P,  will 
be  represented  by  PQ;  while  the 
whole  cylinder  will  contain  as  many 
units  of  volume  as  there  are  units  of 
force  in  the  whole  stress. 

Fig.  1 66  represents    a   uniformly 
varying  stress.     Here,  again,  AB  is 
the  surface  of  action,  and  the  stress  /Y 
is  supposed  to  vary  at  a  uniform  rate 
from  the  axis  OY. 


FIG.  165. 

The  upper  bounding  surface  of  the  cylin- 


FIG.  166. 


drical  figure  which  represents  the  stress 
becomes  a  plane  inclined  to  the  XOY 
plane,  and  containing  the  axis  OY. 

In  this  case,  if   a   represent  the   in- 
tensity of  the  stress  at  a  unit's  distance 
'°  from  OY,  the  stress  at  a  distance  x  from 
OY  will  be/  =  ax>  and  the  total  amount 
of  the  stress  will  be 

P  =  ffpdxdy  =  affxdxdy. 


When  a  stress  is  oblique  to  the  surface  of  action,  it  may  be 
represented  correctly  in  all  particulars,  except  in  direction,  in 
the  above-stated  way. 

§  180.  Centre  of  Stress.  —  The  centre  of  stress,  or  the 
point  of  the  surface  at  which  the  resultant  of  the  stress  acts, 
often  becomes  a  matter  of  practical  importance.  If,  for  con- 
venience, we  employ  a  system  of  rectangular  co-ordinate  axes, 
of  which  the  axes  OX  and  OY  are  taken  in  the  plane  of  the 
surface  on  which  the  stress  acts,  and  if  we  let  /  =  <£(#,  y)  be 


264  APPLIED   MECHANICS. 

the  intensity  of  the  stress  at  the  point  (x,  y)t  we  shall  have, 
for  the  co-ordinates  of  the  centre  of  stress, 

_  ffxpdxdy  _  ffypdxdy 

~~   ffpdxdy*    Jl  ==   ffpdxdy* 

(see  §  42),  where  the  denominator,  or  ffpdxdy,  represents  the 
total  amount  of  the  stress. 

When  the  stress  is  positive  and  negative  at  different  parts 
of  the  surface,  as  in  Fig.  164,  the  case  may  arise  when  the  posi- 
tive and  negative  parts  balance  each  other,  and  hence  the 
stress  on  the  surface  constitutes  a  statical  couple.  In  that  case 

ffpdxdy  =  o. 

§  181.  Uniform  Stress.  —  In  the  case  of  uniform  stress,  we 
have  — 

i°.  The  intensity  of  the  stress  is  constant,  or  /  =  a,  con- 
stant. 

2°.  The  volume  which  represents  it  graphically  becomes  a 
cylinder  with  parallel  and  equal  bases,  as  in  Fig.  165. 

3°.  The  centre  of  stress  is  at  the  centre  of  gravity  of  the 
surface  of  action  ;  for  the  formulae  become,  when  /  is  constant, 

_  pffxdxdy  _  ffxdxdy  _ 
~~   pffdxdy  =  ''  ~  = 


_  pffydxdy  _  ffydxdy  _ 
~~  pffdxdy  ==    ffdxdy   ~  y°' 

where  xot  yw  are  the  co-ordinates  of  the  centre  of  gravity  of  the 
surface. 

Examples  of  uniform  stress  have  already  been  given  in  the 
cases  of  direct  tension,  direct  compression,  and,  in  the  case  of 
riveted  joints,  for  the  shearing-force  on  the  rivet. 


UNIFORMLY   VARYING  STRESS.  265 

§182.  Uniformly  Varying  Stress.  —  Uniformly  varying 
stress  has  already  been  defined  as  a  stress  whose  intensity  varies 
uniformly  from  a  given  line  in  its  own  plane  ;  and  this  line  will 
be  called  the  Neutral  Axis.  Thus,  if  the  plane  be  taken  as  the 
XOY  plane  (Fig.  166),  and  the  given  line  be  taken  as  OY,  we 
shall  have,  if  a  denotes  the  intensity  of  the  stress  at  a  unit's 
distance  from  OY,  and  x  the  distance  of  any  special  point  from 
O  Y,  that  the  intensity  of  the  stress  at  the  point  will  be 

/  =  ace. 

The  total  amount  of  the  stress  will  be 

P  =  affxdxdy. 

The  total  moment  of  the  stress  about  O  Y  will  be  found  by 
multiplying  each  elementary  stress  by  its  leverage.  This  lever- 
age is,  in  the  case  of  normal  stress,  x  ;  hence  in  that  case  the 
moment  of  any  single  elementary  force  will  be 


and  the  total  moment  of  the  stress  will  be 
M  -  affx^dxdy  =  al. 

In  the  case  of  oblique  stress,  this  result  has  to  be  modified, 
as  the  leverage  is  no  longer  x.  Confining  ourselves  to  stress 
normal  to  the  plane  of  action,  we  have,  for  the  co-ordinates  of 
the  centre  of  stress, 

_  ffpxdxdy       affx2dxdy        ffx2dxdy        ffx2dxdy 
''    ffpdxdy   ~~  P  =  ffxdxdy    ~~          xQA 

_  ffpydxdy  _  affxydxdy  __  ffxydxdy  _  ffxydxdy 
~~  ffpdxdy  =  P  ~~   ffxdxdy   ~~         x^A 

since 

P  =  affxdxdy  =  ax0A, 

where  xw  y0,  are  the  co-ordinates  of  the  centre  of  gravity,  and 
A  is  the  area  of  the  surface  of  action. 


266  APPLIED   MECHANICS. 


§  183.  Case  of  a  Uniformly  Varying  Stress  which 
amounts  to  a  Statical  Couple.  —  Whenever  P  ^=  o,  we  have 

affxdxdy  =  o       /.    ffxdxdy  —  o       .-.    x0A  —  o       /.    x0  =  o. 

In  this  case,  therefore,  we  have  — 

i°.  There  is  no  resultant  stress,  and  hence  the  whole  stress 
amounts  to  a  statical  couple. 

2°.  Since  XQ  =  o,  the  centre  of  gravity  of  the  surface  of 
action  is  on  the  axis  <9F,  which  is  the  neutral  axis. 

Hence  follows  the  proposition  :  — 

When  a  uniformly  varying  stress  amounts  to  a  statical  couple, 
the  neutral  axis  contains  (passes  through)  the  centre  of  gravity 
of  the  surface  of  action. 

In  this  case  there  is  no  single  resultant  of  the  stress  ;  but 
the  moment  of  the  couple  will  be,  as  has  been  already  shown, 

M=  a/fx^dxdy. 

§  184.   Example  of  Uniformly  Varying  Stress One  of 

the  most  common  examples  of  uniformly  varying  stress  is  that 
of  the  pressure  of  water  upon  the  sides  of  the  vessel  contain- 
ing it. 

Thus,  let  Fig.  167  represent  the  vertical  cross-section  of  a 
reservoir  wall,  the  water  pressing  against  the 
vertical  face  AB.  It  is  a  fact  established  by 
experiment,  that  the  intensity  of  the  pressure 
of  any  body  of  water  at  any  point  is  propor- 
tional to  the  depth  of  the  point  below  the 
free  upper  level  of  the  water,  and  normal  to 
the  surface  pressed  upon.  Hence,  if  we  sup- 
pose the  free  upper  level  of  the  water  to  be 
even  with  the  top  of  the  wall,  the  intensity 
of  the  pressure  there  will  be  zero  ;  and  if  we  represent  by  CB 
the  intensity  of  the  pressure  at  the  bottom,  then,  joining  A  and 


STRESSES  IN  BEAMS   UNDER    TRANSVERSE  LOAD.      267 


C  we  shall  have  the  intensity  of  the  pressure  at  any  point,  as 
D,  represented  by  ED,  where 

ED  :  CB  =  AD  :  AB. 

Here,  then,  we  have  a  case  of  uniformly  varying  stress  nor- 
mal to  the  surface  on  which  it  acts. 

§  185.  Fundamental  Principles  of  the  Common  Theory 
of  the  Stresses  in  Beams  under 
a    Transverse    Load.  —  Fig.   168 
shows  a  beam  fixed  at  one  end  and 
loaded  at  the  other,  while  Fig.  169 
shows  a  beam    supported   at   the 
ends    and    loaded   at    the    middle. 
Let,  in  each  case,  the  plane  of  the 
paper    contain    a    vertical    longi- 
tudinal section  of  the  beam.     In 
.o      Fig.    168, 
J\        it  is   evi- 
//        dent  that 
/I         the  upper 

/  /         fibres  are  lengthened,  while  the  lower 
/  I          ones  are  shortened,  and  vice  versa  in 
I  I  Fig.    169.      In    either   case,    there    is, 

/  I  somewhere    between    the    tipper   and 

lower  fibres,  a  fibre  which  is  neither 
elongated  nor  com- 
pressed. 

Let    CN    repre- 
sent that  fibre,  Fig. 

1 68,  and    CP,    Fig. 

169.  This  line  may 
be  called  the  neutral 
line  of    the   longitu- 
dinal section  ;  and,  if  a  section  be  made  at  any  point  at  right 


FIG.  169. 


268  APPLIED   MECHANICS. 

angles  to  this  line,  the  horizontal  line  which  lies  in  the  cross- 
section,  and  cuts  the  neutral  lines  of  all  the  longitudinal  sec- 
tions, or,  in  other  words,  the  locus  of  the  points  where  the 
neutral  lines  of  the  longitudinal  sections  cut  the  cross-section, 
is  called  the  Neutral  Axis  of  the  cross-section.  In  the  ordinary 
theory  of  the  stresses  in  beams,  a  number  of  assumptions  are 
made,  which  will  now  be  enumerated. 

ASSUMPTIONS    MADE    IN    THE    COMMON    THEORY    OF    BEAMS. 

ASSUMPTION  No.  i.  —  If,  when  a  beam  is  not  loaded,  a 
plane  cross-section  be  made,  this  cross-section  will  still  be  a 
plane  after  the  load  is  put  on,  and  bending  takes  place.  From 
this  assumption,  we  deduce,  as  a  consequence,  that,  if  a  certain 
cross-section  be  assumed,  the  elongation  or  shortening  per  unit 
of  length  of  any  fibre  at  the  point  where  it  cuts  this  cross-sec- 
tion, is  proportional  to  the  distance  of  the  fibre  from  the  neutral 
axis  of  the  cross-section. 

Proof.  —  Imagine  two  originally  parallel  cross-sections  so 
near  to  each  other  that  the  curve  in  which  that  part  of  the 
neutral  line  between  them  bends  may,  without  appreciable  error, 
be  accounted  circular.  Let  ED  and  GH  (Fig.  168  or  Fig.  169) 
be  the  lines  in  which  these  cross-sections  cut  the  plane  of  the 
paper,  and  let  O  be  the  point  of  intersection  of  the  lines  ED 
and  GH.  Let  OF  —  r,  FL  —  y,  FK  —  /,  LM  =  I  +  a/,  in 
which  a  is  the  strain  or  elongation  per  unit  of  length  of  a  fibre 
at  a  distance  y  from  the  neutral  line,  y  being  a  variable  ;  then, 
because  FK  and  LM  are  concentric  arcs  subtending  the  same 
angle  at  the  centre,  we  shall  have  the  proportion 


r  +  y       I  +  a/  y 

—  =  — —      or      I   +  a  =   I   +  -, 


y 

a  =  —  or      a 

r 


&•• 


ASSUMPTIONS  IN  THE   COMMON  THEORY  OF  BEAMS.    269 

but  as  y  varies  for  different  points  in  any  given  cross-section, 
while  r  remains  the  same  for  the  same  section,  it  follows,  that, 
if  a  certain  cross-section  be  assumed,  the  strain  of  any  fibre  at 
the  point  where  it  cuts  this  cross-section  is  proportional  directly 
to  the  distance  of  this  fibre  from  the  neutral  axis  of  the  cross- 
section. 

ASSUMPTION  No.  2.  —  This  assumption  is  that  commonly 
known  as  Hookes  Law.  It  is  as  follows  :  "  Ut  tensio  sic  vis  ;  " 
i.e.,  The  stress  is  proportional  to  the  strain,  or  to  the  elonga- 
tion or  compression  per  unit  of  length.  As  to  the  evidence  in 
favor  of  this  law,  experiment  shows,  that,  as  long  as  the  mate- 
rial is  not  strained  beyond  safe  limits,  this  law  holds.  Hence, 
making  these  two  assumptions,  we  shall  have  :  At  a  given 
cross-section  of  a  loaded  beam,  tJie  direct  stress  on  any  fibre 
varies  directly  as  the  distance  of  the  fibre  from  the  neutral  axis. 
Hence  it  is  a  uniformly  varying  stress,  and  we  may  repre- 
sent it  graphically  as  follows :  Let 
ABCDj  Fig.  170,  be  the  cross-sec- 
tion of  a  beam,  and  KL  the  neutral 
axis.  Assume  thts  for  axis  OY,  and 
draw  the  other  two  axes,  as  in  the 
figure.  If,  now,  EA  be  drawn  to 
represent  the  intensity  of  the  direct 
(normal)  stress  at  A,  then  will  the 
pair  of  wedges  AEFBKL  and  FlG- I7°- 

DCHGKL  represent  the  stress  graphically,  since  it  is  uni- 
formly varying. 


POSITION   OF   NEUTRAL  AXIS. 


ASSUMPTION  No.  3.  —  It  will  next  be  shown,  that,  on  the 
two  assumptions  made  above,  and  from  the  further  assumption 
that  the  only  resistances  opposed  to  the  bending  of  the  beam 


2 7°  APPLIED   MECHANICS. 

are  the  direct  tensions  and  compressions  of  the  fibres,  it  fol- 
lows that  the  neutral  axis  must  pass  through  the  centre  of 
gravity  of  the  cross-section. 


^^A                   Y 

N 

O 

1    0 

R 

r  E 

AA 

IN      E                     "~  B 

I  w  (  |w| 

FIG.  171.  FIG.  172. 

Since  the  curvatures  in  Figs.  168  and  169  are  exaggerated 
in  order  to  render  them  visible,  Figs.  171  and  172  have  been 
drawn.  If,  now,  we  assume  a  section  DE,  such  that  AD  =  x 
(Fig.  171)  and  NE  =•  x  (Fig.  172),  and  consider  all  the  forces 
acting  on  that  part  of  the  beam  which  lies  to  the  right  of  DE 
(i.e.,  both  the  external  forces  and  the  stresses  which  the  other 
parts  of  the  beam  exert  on  this  part),  we  must  find  them  in 
equilibrium.  The  external  forces  are,  in  Fig.  172,  —  . 

i°.  The  loads  acting  between  B  and  E ;  in  this  case  there 
are  none. 

2°.  The  supporting  force  at  B ;  in  this  case  it  is  equal  to 

— ,  and  acts  vertically  upwards. 

In  Fig.  171  they  are, — 

The  loads  between  D  and  N ;  in  this  case  there  is  only  the 
one,  W  at  N. 

The  internal  forces  are  merely  the  stresses  exerted  by  the 
other  parts  of  the  beam  on  this  part :  they  are,  — 

i°.  The  resistance  to  shearing  at  the  section,  which  is  a 
vertical  stress. 

2°.  The  direct  stresses,  which  are  horizontal. 

Now,  since  the  part  of  the  beam  to  the  right  of  DE  is  at 
rest,  the  forces  acting  on  it  must  be  in  equilibrium  ;  and,  since 


POSITION  OF  NEUTRAL   AXIS.  27 l 

they  are  all  parallel  to  the  plane  of  the  paper,  we  must  have 
the  three  following  conditions  ;  viz.,  — 

i°.  The  algebraic  sum  of  the  vertical  forces  must  be  zero. 

2°.  The  algebraic  sum  of  the  horizontal  forces  must  be  zero. 

3°.  The  algebraic  sum  of  the  moments  of  the  forces  about 
any  axis  perpendicular  to  the  plane  of  the  paper  must  be 
zero. 

But,  on  the  above  assumptions,  the  only  horizontal  forces 
are  the  direct  stresses  :  hence  the  algebraic  sum  of  these  direct 
stresses  must  be  zero ;  or,  in  other  words,  the  direct  stresses 
must  be  equivalent  to  a  statical  couple. 

Now,  it  has  already  been  shown,  that,  whenever  a  uniformly 
varying  stress  amounts  to  a  statical  couple,  the  neutral  axis 
must  pass  through  the  centre  of  gravity  of  the  surface  acted 
upon.  Hence  in  a  loaded  beam,  if  the  three  preceding  assump- 
tions be  made,  it  follows  that  the  neutral  axis  of  any  cross- 
section  must  contain  the  centre  of  gravity  of  that  section. 

By  way  of  experimental  proof  of  this  conclusion,  Barlow 
has  shown  by  experiment,  that,  in  a  cast-iron  beam  of  rectangu- 
lar section,  the  neutral  axis  does  pass  through  the  centre  of 
gravity  of  the  section. 

RESUME. 

The  conclusions  arrived  at  from  the  foregoing  are  as  fol- 
lows :  — 

i°.  That  at  any  section  of  a  loaded  beam,  if  a  horizontal 
line  be  drawn  through  the  centre  of  gravity  of  the  section, 
then  the  fibres  lying  along  this  line  will  be  subjected  neither 
to  tension  nor  to  compression  ;  in  other  words,  this  line  will  be 
the  neutral  axis  of  the  section. 

2°.  The  fibres  on  one  side  of  this  line  will  be  subjected  to 
tension,  those  on  the  other  side  being  subjected  to  compres- 
sion ;  the  tension  or  compression  of  any  one  fibre  being  propor- 
tional to  its  distance  from  the  neutral  axis. 


2/2  APPLIED   MECHANICS. 

§  1 86.  Shearing-Force  and  Bending-Moment In  deter- 
mining the  strength  of  a  beam,  or  the  proper  dimensions  of  a 
beam  to  bear  a  certain  load,  when  we  assume  the  neutral  axis 
to  pass  through  the  centre  of  gravity  of  the  cross-section,  we 
have  imposed  the  second  of  the  three  last-mentioned  conditions 
of  equilibrium.  The  remaining  two  conditions  may  otherwise 
be  stated  as  follows  :  — 

i°.  The  total  force  tending  to  cause  that  part  of  the  beam 
that  lies  to  one  side  of  the  section  to  slide  by  the  other  part, 
must  be  balanced  by  the  resistance  of  the  beam  to  shearing  at 
the  section. 

2°.  The  resultant  moment  of  the  external  forces  acting  on 
that  part  of  the  beam  that  lies  to  one  side  of  the  section,  about 
a  horizontal  axis  in  the  plane  of  the  section,  must  be  balanced 
by  the  moment  of  the  couple  formed  by  the  resisting  stresses. 

The  shearing-force  at  any  section  is  tJie  force  with  which  the 
part  of  the  beam  on  one  side  of  the  section  tends  to  slide  by  the 
part  on  the  other  side.  In  a  beam  free  at  one  end,  it  is  equal  to 
the  sum  of  the  loads  between  the  section  and  the  free  end.  In 
a  beam  supported  at  both  ends,  it  is  equal  in  magnitude  to  the 
difference  between  the  supporting  force  at  either  end,  and 
the  sum  of  the  loads  between  the  section  and  that  support. 

The  bending-moment  at  any  section  is  the  resultant  moment 
of  the  external  forces  acting  on  the  part  of  the  beam  to  one 
side  of  the  section,  these  moments  being  taken  about  a  hori- 
zontal axis  in  the  section. 

In  a  beam  free  at  one  end,  it  is  equal  to  the  sum  of  the 
moments  of  the  loads  between  the  section  and  the  free  end, 
about  a  horizontal  axis  in  the  section. 

In  a  beam  supported  at  both  ends,  it  is  the  difference  be- 
tween the  moment  of  either  supporting  force,  and  the  sum  of 
the  moments  of  the  loads  between  the  section  and  that  sup- 
port ;  all  the  moments  being  taken  about  a  horizontal  axis  in 
the  section. 


SHEARING-FORCE   AND   BENDING-MOMENT.  273 

Hence  the  two  conditions  of  equilibrium  may  be  more 
briefly  stated  as  follows  :  — 

i°.  The  shearing-force  at  the  section  must  be  balanced 
by  the  resistance  opposed  by  the  beam  to  shearing  at  the 
section. 

2°.  The  bending-moment  at  the  section  must  be  balanced 
by  the  moment  of  the  couple  formed  by  the  resisting  stresses. 

It  is  necessary,  therefore,  in  determining  the  strength  of  a 
beam,  to  be  able  to  determine  the  shearing-force  and  bending- 
moment  at  any  point,  and  also  the  greatest  shearing-force  and 
the  greatest  bending-moment,  whatever  be  the  loads. 

A  table  of  these  values  for  a  number  of  ordinary  cases  will 
now  be  given  ;  but  I  should  recommend  that  the  table  be  merely 
considered  as  a  set  of  examples,  and  that  the  rules  already 
given  for  finding  them  be  followed  in  each  individual  case. 

Let,  in  each  case,  the  length  of  the  beam  be  /,  and  the 
total  load  W.  When  the  beam  is  fixed  at  one  end  and  free  at 
the  other,  let  the  origin  be  taken  at  the  fixed  end ;  when  it  is 
supported  at  both  ends,  let  it  be  taken  directly  over  one  support. 
Let  x  be  the  distance  of  any  section  from  the  origin.  Then  we 
shall  have  the  results  given  in  the  following  table :  — 


2/4 


APPLIED   MECHANICS. 


At  Dista 
from  O 


I  i 


51 


S-2 


4'i 


l? 

-§1 


11 


s  t-j 


*.  o 


c  <u 

rt  i- 
<u  ^* 
W 


MOMENTS   OF  INERTIA    OF  SECTIONS.  2/5 

In  a  beam  fixed  at  one  end  and  free  at  the  other,  the  great- 
est shearing-force,  and  also  the  greatest  bending-moment,  are  at 
the  fixed  end.  In  a  beam  supported  at  both  ends,  and  loaded 
at  the  middle,  or  with  a  uniformly  distributed  load,  the  greatest 
shearing-force  is  at  either  support,  the  greatest  bending-moment 
being  at  the  middle.  In  the  last  case  (i.e.,  that  of  a  beam  sup- 
ported at  the  ends,  and  having' a  single  load  not  at  the  middle), 
the  greatest  bending-moment  is  at  the  load ;  the  greatest  shear- 
ing-force being  at  that  support  where  the  supporting  force  is 
greatest. 

§  187.  Moments  of  Inertia  of  Sections.  —  In  the  usual 
methods  of  determining  the  strength  of  a  beam  or  column,  it 
is  necessary  to  know,  i°,  the  distance  from  the  neutral  axis  of 
the  section  to  the  most  strained  fibres;  2°,  the  moment  of  in- 
ertia of  the .  section  about  the  neutral  axis.  The  manner  of 
finding  the  moments  of  inertia  has  been  explained  in  Chap.  II. 

In  the  following  table  are  given  the  areas  of  a  large  number 
of  sections,  and  also  their  moments  of  inertia  about  the  neutral 
axis,  which  is  the  axis  YY  in  each  case.  These  results  should 
be  deduced  by  the  student. 


276 


APPLIED  MECHANICS. 


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APPLIED   MECHANICS. 


§  1 88.  Cross-Sections  of  Phoenix  Columns  considered 
as  made  of  Lines.  —  It  is  to  be  observed  that  the  moments 
of  inertia  are  the  same  for  all  axes  passing  through  the  centre. 
Thickness  —  /,  radius  of  round  ones  —  ry  area  of  each  flange 
=  a,  length  of  each  flange  =  /. 


Figure. 


Description. 


A. 


Four  flanges 


27ir/  +  40 


20 


Eight  flanges 


Square,  four  flanges, 
r  =  radius  of  cir- 
cumscribed circle 


214 


Six  flanges 


mrt  +  6a 


nr3*  +  y 


REPRESENTATION  OF  B ENDING-MOMENTS. 


287 


§  189.   Graphical  Representation  of  Bending-Moments. — 

The  bending-moment  at  each  point  of  a  loaded  beam  may  be 
represented  graphically  by  lines  laid  off  to  scale,  as  will  be 
shown  by  examples. 

I.  Suppose   we   have    the    cantilever   shown   in    Fig.  215, 
loaded  at  D  with  a  load  W :   then 

will    the    bending-moment    at    any 

section,    as   at  Ft    be   obtained   by 

multiplying  W  by  FD ;  that  at  AC 

being  W  X  (AB).     If,  now,  we  lay 

off  CE  to  scale  to  represent  this, 

i.e.,  having  as  many  units  of  length 

as  there  are  units  of  moment  in  the  product  W  X  (AB\  and 

join  E  with  D,  then  will  the  ordinate  FG  of  any  point,  as  G, 

represent  (to  the  same  scale)  the  bending-moment  at  a  section 

through  F. 

II.  If  we  have  a  uniformly  distributed  load,  we  should  have, 
for  the  line  corresponding  to  CE  in  Fig.  215,  a  curve.     This  is 

shown  in  Fig.  216,  where  we  have  the 
uniformly  distributed  load  EIGF.  If 
we  take  the  origin  at  D,  as  before, 
we  have,  for  the  bending-moment,  at  a 
distance  x  from  the  origin,  as  has  been 


FIG.  215. 


FIG.  216. 


W 
shown,  — (/  — 


and  by  giving  x  dif- 


ferent values,  and  laying  off  the  corresponding  value  of  the 
bending-moment,  we  obtain  the  curve  CA,  any  ordinate  of 
which  will  represent  the  bending-moment  at  the  corresponding 
point  of  the  beam. 

When  we  have  more  than  one  load  on  a  beam,  we  must  draw 
the  curve  of  bending-moments  for  each  load  separately,  and 
then  find  the  actual  bending-moment  at  any  point  of  the  beam 


288 


APPLIED   MECHANICS. 


by. taking  the  sum  of  the  ordinates  (drawn  from  that  point)  of 
each  of  these  separate  curves  or  straight  lines.  If  we  then 
draw  a  new  curve,  whose  ordinates  are  these  sums,  we  shall  have 
the  actual  curve  of  bending-moments  for  the  beam  as  loaded. 
Some  examples  will  now  be  given,  whicli  will  explain  them- 
selves. 

III.  Fig.  217  shows  a  cantilever  with  three  concentrated 
loads.  The  line  of  bending-moments 
for  the  load  at  C  is  CEy  that  for  the 
load  at  O  is  OF,  and  for  the  load  at  P 
is  PG.  They  are  combined  above  the 
beam  by  laying  off  AH  =  DE,  HK  = 
DF,  and  KL  =  DG,  and  thus  obtaining 
the  broken  line  LMNB,  which  is  the 
line  of  bending-moments  of  the  beam 
loaded  with  all  three  loads. 


FIG.  217. 

IV.  Fig.  218  shows  the  case  of  a  beam  supported  at  both 
ends,  and   loaded  at  a   single   point 

D;  ALB  is  the  line  of  bending- 
moments  when  the  weight  of  the 
beam  is  disregarded,  so  that  xy  = 
bending-moment  at  x.  FlG_  2l8_ 

V.  Fig.  219  shows  the  case  of  a  beam  supported  at  the  ends, 

and  loaded  with  three  concentrated 
loads  at  the  points  B,  C,  and  D  re- 
spectively ;  the  lines  of  bending-mo- 
ments for  each  individual  load  being 
respectively  AFE,  AGE,  and  AHE, 
FIG.  219.  and  the  actual  line  of  bending-mo- 

ments being  AKLME. 


REPRESENTATION  OF  BENDING-MOMENTS.  289 

VI.    Fig.  220  shows  the  case  of  a  beam  supported  at  the 
ends,  and  loaded  with  a  uniformly  dis-      A 
tributed   load;    the   line    of    bending- 
moments   being   a    curve,    ACDB,   as 


shown  in  the  figure.  FIG.  220. 

VII.  In  Fig.  221  we  have  the  case  of  a  beam,  over  a  part  of 
which,  viz.,  EFt  there  is  a  distributed  load;    the  rest  of  the 

beam  being  unloaded.  The  line  of 
bending-moments  is  curvilinear  be- 
tween E  and  F,  and  straight  outside 
of  these  limits.  It  isAGSHB;  and, 
when  the  curve  is  plotted,  we  can 
find  the  greatest  bending-moment 

graphically  by  finding  its  greatest  ordinate.  We  can  also 
determine  it  analytically  by  first  determining  the  bending- 
moment  at  a  distance  x  from  the  origin,  and  on  the  side 
towards  the  resultant  of  the  load,  and  then  differentiating. 
This  process  is  shown  in  the  following:  — 
Let  A  (Fig.  222)  be  the  point  where 
the  resultant  of  the  load  acts,  and  O  the 


,  n  A- 

middle  of    the   beam,   and    let  w  be    the  c  OA     B 

load  per  unit  of  length  ;  let  OA  =  a,  AB  = 
A  C  =  b,  and  ED  =  2c,  so  that  the  whole  load  =  2wb :  there- 

a  +  c       wb(a  -f-  c) 
fore  supporting  force  at  D  =  2wb      —  =  -  — . 

If  we  take  a  section  at  a  distance  x  from  O  to  the  right,  we 
shall  have,  for  the  bending-moment  at  that  section, 

wb(a  -f  c)  w 

(f  —  x) (a  -f-  b  —  x}2  =  a  maximum. 

Differentiate,  and  we  have 

+  c)                                                         a(c  —  &) 
L  +  w(a+b-x    =o     .*.    x=— -; 


2QO 


APPLIED  MECHANICS. 


hence  the  greatest  bending-moment  will  be 

—  (c -1 (a  -f-  b  —  a  +  - 

c         \  c      /        2  \ 


=  —(a  +  c)  (c2  —  ac  +  ab)  ---  (a2  -f  zac  +  c2) 

—  —  (a2b  —  2a2c  4- 

2.C2 


—  be2}. 


VIII.  In  Figs.  223  and  224  we  have  the  case  of  a  beam 
supported  at 
the  ends,  and 
loaded  with  a 
uniformly  dis- 
tributed load, 
and  also  with 


FIG.  224. 


FIG.  223. 

is  at  D,  and  in 


a     c  o  n  c  e  n- 
trated     load. 
In    the    first 
figure,   the   greatest  bending-moment 
the  second  at  C. 


IX.  In  Fig.  225  we  have  a  beam  supported  at  A  and  B,  and 
loaded  at   C  and  D  with    equal  B 

weights  ;  the  lengths  ot  AC  and 
.Z?Z>  being  equal.  We  have,  con- 
sequently, between  A  and  B,  a 
uniform  bending-moment ;  while 
on  the  left  of  A  and  on  the  right  FlG- 225' 

of  B  we  have  a  varying  bending-moment.     The  line  of  bending- 
moments  is,  in  this  case,  CabD. 

We  may,  in  a  similar  way,  derive  curves  of  bending-moment 
for  all  cases  of  loading  and  supporting  beams. 


STRESSES  AT  DIFFERENT  PARTS   OF  A   BEAM.          2gi 

§  190.  Mode  of  Procedure  for  Ascertaining  the  Stresses 
at  Different  Parts  of  a  Beam  when  the  Loads  and  the  Di- 
mensions are  given.  —  When  the  dimensions  of  a  beam,  the 
load  and  its  distribution,  and  the  manner  of  supporting  are 
given,  and  it  is  desired  to  find  the  actual  intensity  of  the  stress 
on  any  particular  fibre  at  any  given  cross-section,  we  must  pro- 
ceed as  follows  :  -  - 

i°.  Find  the  actual  bending-moment  (M)  at  that  cross-sec- 
tion. 

2°.  Find  the  moment  of  inertia  (7)  of  the  section  about  its 
neutral  axis. 

3°.  Observe,  that,  from  what  has  already  been  shown,  the 
moment  of  the  couple  formed  by  the  tensions  and  compressions 
is  alj  where  a  —  intensity  of  stress  of  a  fibre  whose  distance 
from  the  neutral  axis  is  unity,  and  that  this  moment  must  equal 
the  bending-moment  at  the  section  in  order  to  secure  equilib- 
rium. Hence  we  must  have 

al  =  M. 

Moreover,  if  /  denote  the  (unknown)  intensity  of  the  stress 
of  the  fibre  where  the  stress  is  desired,  and  if  y  denote  the 
distance  of  this  fibre  from  the  neutral  axis,  we  shall  have 

• 


,     . 

from  which  equation  we  can  determine/. 

EXAMPLES. 

• 

i.  Given  a  beam  18  feet  span,  supported  at  both  ends,  and  loaded 
uniformly  (its  own  weight  included)  with  1000  Ibs.  per  foot  of  length. 
The  cross-section  is  a  T,  where  area  of  flange  =  3  square  inches, 
area  of  web  =  4  square  inches,  height  =  10  inches.  Find  (a)  the 


APPLIED  MECHANICS. 


bending-moment  at  3  feet  from  one  end  ;  (b)  the  greatest  bending- 
moment  ;  (c)  the  greatest  intensity  of  the  tension  at  each  of  the 
above  sections  ;  (d)  the  greatest  intensity  of  the  compression  at  each 
of  these  sections. 

2.  Given  an  I-beam  with  equal  flanges,  area  of  each  flange  =  3 
square  inches,  area  of  web  =  3  square  inches,  height  =  10  inches;  the 
beam  is  12  feet  long,  supported  at  the  ends,  and  loaded  uniformly  (its 
own  weight  included)  with  a  load  of  2000  Ibs.  per  foot  of  length.  Find 
(a)  the  bending-moment  at  a  section  one  foot  from  the  end  ;  (b)  the 
greatest  bending-moment;  (c)  the  greatest  intensity  of  the  stress  at. 
each  of  the  above  cross-sections. 


§  191.  Mode  of  Procedure  for  Ascertaining  the  Dimen- 
sions of  a  Beam  to  bear  a  Certain  Load,  or  the  Load  that 
a  Beam  of  Given  Dimensions  and  Material  is  Capable  of 
Bearing.  —  If  we  wish  to  determine  the  proper  dimensions, 
of  the  beam  when  the  load  and  its  distribution,  as  well  as  the 
manner  of  supporting,  are  given,  so  that  it  shall  nowhere  be 
strained  beyond  safe  limits,  or  if  we  wish  to  determine  the 
greatest  load  consistent  with  safety  when  the  other  quantities, 
are  given,  we  must  impose  the  condition  that  the  greatest 
intensity  of  the  tension  to  which  any  fibre  is  subjected  shall 
not  exceed  the  safe  working-strength  for  tension  of  the  mate- 
rial of  which  the  beam  is  made,  and  the  greatest  intensity  of 
the  compression  to  which  any  fibre  is  subjected  shall  not  exceed 
the  safe  working-strength  of  the  material  for  compression. 

Thus,  we  must  in  this  case  first  determine  where  is  the 
section  of  greatest  bending-moment  (this  determination  some- 
times involves  the  use  of  the  Differential  Calculus). 

Next  we  must  determine  the  magnitude  of  the  greatest 
bending-moment,  absolutely  if  the  load  and  length  of  the  beani 
are  given  (if  not,  in  terms  of  these  quantities),  and  then  equate 
this  to  the  moment  of  the  resisting  couple. 

Thus,  if  M0  is  the  greatest  bending-moment,  70  the  moment 
of  inertia  of  that  section  where  this  greatest  bending-moment 


WORKING-S  TRENG  TH. 


acts,  and  if  ft  =  working-strength  per  square  inch  for  tension, 
fc  =  working-strength  per  square  inch  for  compression,  yt  = 
distance  of  most  stretched  fibre  from  the  neutral  axis,  and  yc 
=  distance  of  most  compressed  fibre  from  the  neutral  axis, 

then  will  —  be  the  greatest  tension  per  square  inch,  at  a  unit's 
distance  from  the  neutral  axis,  consistent  with  safety  against 
tearing,  and  —  the  greatest  compression  per  square  inch,  at  a 

unit's  distance  from  the  neutral  axis,  consistent  with  safety 
against  crushing. 

Of  course  the  least  of  these  must  not  be  exceeded  in  the 
actual  beam.     Hence  we  must  put 


where  -  is  taken  as  the  lesser  of  the  two  quantities  —  and  —  • 

y  yt      yc 


MODULUS   OF   RUPTURE. 

The  modulus  of  rupture  is  the  greatest  tension  or  com- 
pression per  square  inch  to  which  the  most  strained  fibre  of 
the  beam  is  subjected  when  the  beam  is  just  on  the  point 
of  breaking. 

i 

WORKING-STRENGTH. 

The  working-strength  per  square  inch  of  a  material  for 
transverse  strength  is  the  greatest  stress  per  square  inch  to 
which  it  is  safe  to  subject  the  most  strained  fibre  of  the  beam. 
It  is  usually  obtained  by  dividing  the  modulus  of  rupture  by 
some  factor  of  safety,  as  3  or  4. 


294  APPLIED   MECHANICS. 

§  192.  EXAMPLES. 

1.  Given,  as  the  modulus  of  rupture  of  a  spruce  beam,  4000  Ibs.  per 
square  inch  :  find  its  breaking-load,  assuming  it  to  be  4  inches  wide 
and  12  inches  deep,  the  span  being  18  feet,  the  load  being  uniformly 
distributed  over  its  entire  length. 

2.  Suppose  such  a  beam  to  break  with  a  load  at  the  middle  of  5000 
Ibs. :  find  its  modulus  of  rupture. 

3.  Given  a  T-beam  fixed  at  one  end,  and  loaded  uniformly.     The 
area  of  the  flange  is  3  square  inches,  that  of  the  web  also  3  square 
inches,  height  =  10  inches.    The  beam  is  4  feet  long.    Find  the  greatest 
load  it  will  bear  with  safety,  the  working-strength  per  square  inch  of  the 
material  being,  for  tension,  10000  Ibs.,  and  for  compression  8000  Ibs. 

4.  Given  an  I-beam,  area  of  each  flange  being  3  square  inches,  and 
area  of  web  3  square  inches,  height  =  12  inches,  span  8  feet,  supported 
at  the  ends,  and  loaded  uniformly  :  what  load  will  it  bear  with  safety,  the 
working-strength  of  the  iron  for  tension,  and  also  for  compression,  being 
1 2000  Ibs.  per  square  inch. 

5.  Given  a  beam  (Fig.  226)  supported  at  both  ends,  and  loaded, 
i°,  with  w  pounds  per  unit  of  length  uniformly,  and  2°,  with  a  single 
load  W  at  a  distance  a  from  the  left-hand  support :  find  the  position 
of  the  section  of  greatest  bending-moment,  and  the  value  of  the  greatest 
bending-moment. 


Solution. 

i.  Left-hand  supporting-force 


-  a) 


(0  Right-hand  supporting-force 

FIG.  226.  tvl         Wd 

=  T  +  —  ' 

2.  Assume  a  section  at  a  distance  x  from  the  left-hand  support 
(this  support  being  the  origin),  and  the  bending-moment  at  that. 
section  is,  — 


WORKING-STRENGTH. 


when  x  <  a, 

—  a)  \  wx2 


!wl       W(l  —  a)  \          wx2 
T  +  -— "I*-— '' 

and  when  x  >  a, 

I  )  2 

To  find  the  value  of  x  for  the  section  of  greatest  bending-moment, 
differentiate  each,  and  put  the  first  differential  co-efficient  =  zero. 
We  shall  thus  have,  in  the  first  case, 

Wl       W(l  -a)                                       I        W(l  -  a) 
1 -, wx  =  o,  or  x  =  — I ; ; 

2  /  2    ^  Wl 

and  in  the  second  case, 

wl       W(l  -a)  I        W(l  -a)        W 

1 -, wx  —  W  —  o,   or  x  =  — h 7 '—  • — • 

2  /  2  Wl  W 

Now,  whenever  the  first  is  <  #,  or  the  second  is  >  a,  we  shall  have 
in  that  one  the  value  of  x  corresponding  to  the  section  of  greatest 
bending-moment.  But  if  the  first  is  >  a,  and  the  second  <  a,  then  the 
greatest  bending-moment  is  at  the  concentrated  load. 

These  conclusions  will  be  evident  on  drawing  a  diagram 
representing  the  bending-moments  graphically,  as  in  Figs.  223 
and  224 ;  and  the  greatest  bending-moment  may  then  be  found 
by  substituting,  in  the  corresponding  expression  for  the  bend- 
ing-moment, the  deduced  value  of  x. 

6.  Given  an  I-beam  10  feet  long,  supported  at  both  ends,  and 
loaded,  at  a  distance  2  feet  to  the  left  of  the  middle,  with  20000  pounds. 
Find  the  bending-moment  at  the  middle,  the  greatest  bending-moment, 
also  the  greatest  intensity  of  the  tension,  and  that  of  the  compression  at 
each  of  these  sections. 

Given  Area  of  upper  flange  =  8  sq.  in. 

Area  of  lower  flange  =  5  sq.  in. 

Area  of  web  =  7  sq.  in. 

Total  depth  =  14  in. 


296  APPLIED   MECHANICS. 

i 

7.  Given  a  beam  (Fig.  227)  18  feet  long,  loaded  at  A  with  1000 

o       A B c    Ibs.,  and  at  B  with  2000  Ibs. ;  the  beam 

*"!  weighing  200  Ibs.  (OA  =  3  feet,  OB 
—  10  feet).  Find  the  section  of  great- 
est bending-moment,  and  the  bending- 
moment  at  that  section. 

FIG.  227. 

§  193.  Beams  of  Uniform  Strength.  —  Abeam  of  uniform 
strength  (technically  so  called)  is  one  in  which  the  dimensions 
of  the  cross-section  are  varied  in  such  a  manner,  that,  at  each 
cross-section,  the  greatest  intensity  of  the  tension  shall  be 
the  same,  and  so  also  the  greatest  intensity  of  the  com- 
pression. 

Such  beams  are  very  rarely  used ;  and,  as  the  cross-section 
varies  at  different  points,  it  would  be  decidedly  bad  engineering 
to  make  them  of  wood,  for  it  would  be  necessary  to  cut  the 
wood  across  the  grain,  and  this  would  develop  a  tendency  to 
split. 

In  making  them  of  iron,  also,  the  saving  of  iron  would  gen- 
erally be  more  than  offset  by  the  extra  cost  of  rolling  such  a 
beam.  Nevertheless,  we  will  discuss  the  form  of  such  beams  in 
the  case  when  the  section  is  rectangular. 

In  all  cases  we  have  the  general  equation 


applying  at  each  cross-section,  where  M  =.  bending-moment 
(section  at  distance  x  from  origin),  /  =  moment  of  inertia  of 
same  section,  y  =  distance  from  neutral  axis  to  most  strained 
fibre,  and  p  =  intensity  of  stress  on  most  strained  fibre;  the 
condition  for  this  case  being  that  /  is  a  constant  for  all  values 
of  x  (i.e.,  for  air  positions  of  the  section),  while  M,  7,  and  y 
are  functions  of  x. 


BEAMS  OF  UNIFORM  STRENGTH. 


As  we  are  limiting  ourselves  to  rectangular  sections,  if  we 
let  b  =  breadth  and  h  =  depth  of  rectangle  (one  or  both  vary- 
ing with  x),  we  shall  have 

M  =  t  bh* 
6 

as  the  condition  for  such  a  beam,  with  /  a  constant  for  all  values 
of  x,  when  the  same  load  remains  on  the  beam. 

We  must,  therefore,  have  bkz  proportional  to  M.  Hence, 
assuming  the  origin  as  before, 

i°.  Fixed  at  one  end,  load  at  the  other,          bh2  =(-)  W(l  -  x). 
2°.  Fixed  at  one  end,  uniformly  loaded,          bh2  =  (-  —  )  (/-  x)2. 

\^    2/' 

{/  /6  W\ 

2  \p     2  / 

for*>£,^=(^)(/-*). 
2  \/     2  / 

4°.  Supported  at  ends,  uniformly  loaded,       bh2  ==(  --  )(lx  —  x2). 

\P   2/' 

Now,  this  variation  of  section  may  be  accomplished  in  one 
of  two  ways:  ist,  by  making  k  constant,  and  letting  b  vary; 
and  2d,  by  making  b  constant,  and  letting  h  vary.  Thus,  in 
the  first  case  above  mentioned,  if  h  is  constant,  we  have,  for  the 
plan  of  the  beam, 


and  if  one  side  be  taken  parallel  to  the  axis  of  the  beam,  this 
will  be  the  equation  of  the  other  side  ;  and,  as  this  is  the  equa- 
tion of  a  straight  line,  the  plan  will  be  a  triangle. 


298  APPLIED   MECHANICS. 

If,  on  the  other  hand,  b  be  constant,  and  h  vary,  we  shall 
have,  for  the  vertical  longitudinal  section  of  the  beam, 


and,  if  one  side  be  taken  as  a  straight  line  in  the  direction  of 
the  axis,  the  other  will  be  a  parabola. 

A  similar  reasoning  will  give  the  plan  or  elevation  respect- 
ively in  each  case ;  and  these  can  be  readily  plotted  from  their 
equations. 

CROSS-SECTION   OF  EQUAL  STRENGTH. 

A  cross-section  of  equal  strength  (technically  so  called)  is- 

one  so  proportioned  that  the  greatest  intensity  of  the  tension 

shall  bear  the  same  ratio  to  the  breaking  tensile  strength  of  the 

material  as  the  greatest  intensity  of  the  compression  bears  to 

the  breaking  compressive  strength   of   the  material.     This  is 

accomplished,  as  will  be  shown  directly,  by  so  arranging  the 

form  and  dimensions  of  the  section  that  the  distance  of  the 

neutral  axis  from  the  most  stretched  fibre  shall  bear  to    its 

distance  from  the  most  compressed  fibre  the  same  ratio  that 

the  tensile  bears  to  the  compressive  strength  of  the  material. 

Let  fc  =  breaking-strength  per  square  inch  for  compression,, 

ft  =  breaking-strength  per  square  inch  for  tension, 

yc  =  distance  of   neutral  axis  from  most   compressed 

fibre, 

yt  =  distance  of  neutral  axis  from  most  stretched  fibre. 

If  pc  =  actual  greatest  intensity  of  compression,  and  pt  = 

actual  greatest  intensity  of   tension,  then,  for  a  cross-section 

of  equal  strength,  we  must  have,  according  to  the  definition, 

— =  — ;  but  we  have  —  =  —  =  intensity  of  stress  at  a  unit's 
Pt     ft  yc       yt 


CROSS-SECTION  OF  EQUAL   STRENGTH.  2Q9 

distance  from  the  neutral  axis.     Hence,  combining  these  two, 
we  obtain 


EXAMPLE. 

Suppose  we  have/-  =  80000  Ibs.  per  square  inch,  and/  =  20000 
Ibs.  per  square  inch.  :  find  the  proper  proportion  between  the  flange  Al 
and  the  web  A2  of  a  T-section  whose  depth  is  h. 

§  194.  Deflection  of  Beams.  —  We  have  already  seen  (§  185),. 
that,  in  the  case  of  a  beam  which  is  bent  by  a  transverse  load, 
we  have 


where  (having  assumed  a  certain  cross-section  whose  distance 
from  the  origin  is  x)  a  =  the  strain  of  a  fibre  whose  distance 
from  the  neutral  axis  is  y,  and  r  =  radius  of  curvature  of 
the  neutral  lamina  at  the  section  in  question.  Hence  follows  the 
equation 


but  from  the  definition  of  E,  the  modulus  of  elasticity,  we  shall 
have 


where  /  =  intensity  of  the  stress  at  a  distance  y  from  the 
neutral  axis. 

Hence  it  follows,  assuming  Hooke's  law,  that 

£  =  JL  =  Ll 

r       Ey       E  y 

We  have  already  seen,  that,  disregarding  signs,  M  =  -  / 


3OO  APPLIED  MECHANICS. 

(making,  of  course,  the  two  assumptions  already  spoken  of 
when  this  formula  was  deduced),  where  M  =  bending-moment 
at,  and  /  =  moment  of  inertia  of,  the  section  in  question  ;  i.e., 
of  that  section  whose  distance  from  the  origin  is  x.  This  gives 

f-  = ,  if,  denoting  tension  by  the  -f-  sign,  and  taking  y 

y         ! 

positive  upwards,  we  call  M  positive  when  it  tends  to  cause 
tension  on  the  lower,  and  compression  on  the  upper,  side ;  these 
being  the  conventions  in  regard  to  signs  which  we  shall  adopt 
in  future.  Hence,  by  substitution,  we  have 

I  =   P-  =  -  *L.  (i) 

r       Ey  El 

Now,  if  we  assume  the  axis  of  x  coincident  with  the  neutral 
line  of  the  central  longitudinal  section  of  the  beam,  and  the 
axis  of  v  at  right  angles  to  this,  and  v  positive  upwards,  no 
matter  where  the  origin  is  taken,  we  shall  always  have,  as  is 
shown  in  the  Differential  Calculus, 


£  _  dx2 

r 


Hence  equation  (i)  becomes 

dx2  M 


M  and  /  being  functions  of  x :  and,  when  we  can  integrate 
this  equation,  we  can  obtain  v  in  terms  of  x,  thus  having  the 
-equation  of  the  elastic  curve  of  the  neutral  line ;  and,  by  com- 
puting the  value  of  v  corresponding  to  any  assumed  value  of  x, 
we  can  obtain  the  deflection  at  that  point  of  the  beam. 


FORMULA  FOR  SLOPE   AND  DEFLECTION.  3OI 

The  above  equation  (2)  is,  as  a  rule,  too  complicated  to  be 
integrated,  except  by  approximation ;  and  the  approximation 
usually  made  is  the  following  :  — 

Since  in  a  beam  not  too  heavily  loaded,  the  slope,  and  con- 
sequently the  tangent  of  the  slope  (or  angle  the  neutral  line 
makes  with  the  horizontal  at  any  point),  is  necessarily  small,  it 

follows  that  —  is  very  small,  and  hence  (— )  is  also  very  small, 
dx  \dxl 

and  i-f-(—)   is  nearly  equal  to  unity.     Making  this  substitu- 
tion, we  obtain,  in  place  of  equation  (2), 


and  this  is  the  equation  with  which  we  always  start  in  com- 
puting the  slope  and  deflection  of  a  loaded  beam,  or  in  finding 
the  equation  of  the  elastic  line. 

By  one  integration  (suitably  determining  the  arbitrary  con- 
stant) we  obtain  the  slope  whose  tangent  is  — ,  and  by  a  second 

integration  we  obtain  the  deflection  v  at  a  distance  x  from  the 
origin ;  and  thus,  by  substituting  any  desired  value  for  x,  we 
can  obtain  the  deflection  at  any  point. 

§  195.  Ordinary   Formulae   for    Slope   and    Deflection. — 
We  may  therefore  write,  if  i  is  the  circular  measure  of  the 

slope  at  a  distance  x  from  the  origin,  since  i  =  tan  i  =  — 

nearly, 

d2v  _  M 

~d^~~~^r 
<•  =  ?  = 

dx 


3O2  APPLIED   MECHANICS. 

In  these  equations,  of  course,  E  is  taken  as  a  constant,  M 
must  ALWAYS  be  expressed  in  terms  of  x,  and  so  also  must  / 
whenever  the  section  varies  at  different  points.  When,  how- 
ever, the  section  is  uniform,  /  is  constant,  and  the  formulae 
reduce  to  !  - 


§  196.  Special  Cases.  —  i°.  Let  us  take  a  cantilever  loaded 
with  a  single  load  at  the  free  end.  Assume  the  origin,  as 
before,  at  the  fixed  end,  and  let  the  beam  be  one  of  uniform 
section.  We  then  have  M  =  —W(l  —  x\ 


To  determine  c,  observe  that  when  x  =  o,  i  —  o ; 

/.    c  =  o  .*.    i  = j&  —  — ) 

EJ\          2  / 

is  the  slope  at  a  distance  ;tr  from  the  origin. 
The  deflection  at  the  same  point  will  be 


/W 
-EI 


but  when  x  =  o,  v  =  o        .'.     c  =  o        .'.     the  deflection  at 
a  distance  x  from  the  origin  will  be 

W  /lx*       *3\  ,  v 

V  =    --  1  -    —    —I-  (2) 

EI\  2         6  y1 

The  equations  (i)  and  (2)  give  us  the  means  of  finding  the 
slope  and  deflection  at  any  point  of  the  beam. 

To  find  the  greatest  slope  and  deflection,  we  have  that  both 
expressions  are  greatest  when  x  —  /.  Hence,  if  i0  and  v0  rep- 
resent the  greatest  slope  and  deflection  respectively, 

Wl2 


SPECIAL    CASES.  3O3 


2°.  Next  take  the  case  of  a  beam  supported  at  both  ends 
and  loaded  uniformly,  the  load  per  unit  of  length  being  w. 
Assume  the  origin  at  the  left-hand  end  ;  then 

wl        wx2       w 
M  =  — x =  —  (lx  —  x2}         and         W  •=.  wl 

2  2  2  V  ' 


W 


/ 
(fc_ 


To  determine  c,  we  have  that  when  x  =  -,  then  i  =  o ; 

W      //3  /3  \  7C//3 

24/  2^EI 

'lx2       x*\         wl*  w  » 

-  4^?3  —   /3)       (!) 


2  3  /  2^EI          2^EI 

/w    r 
>dx  =  — — _  I  (6lx2  —  AX*  - 
^EIJ 

But  when  ^  =  0,^  =  0; 

/.    c  —  o 


w 


For  the  greatest  slope,  we  have  x  =  o,  or  x  =  I; 


= 


24^7 


For  the  greatest  deflection,  x  —  -  ; 


—  w  5 

=  384^7 


304  APPLIED   MECHANICS. 

3°.  Take  the  case  of  a  beam  supported  at  both  ends,  and 
loaded  at  the  middle  with  a  load  W. 

Assume,  as  before,  the  origin  at  the  left-hand  support.. 
Then  we  shall  have 

W  I  W  I 

M  =  —x,      x  <  -,     and     M  =  —  (/  —  x)  when  x  >  — 

Therefore,  for  the  slope  up  to  the  middle,  we  have 

WC  W  x2 

i  =  ZBJJ  xdx  =  IE/7  +  '' 


When  x  =  —  ,  then  i  =  o  ; 


wr 


(I) 

i+J-sJ.      \  q.f 

and 

W    rf  a      /2U          JF  /*3      l*x\    . 

v  =  —p-T  I  (x* }dx  —  —^rX 1  +  c. 

4^,yt/     \  47  4^iy  \$          47 

But  when  ^-  =  0,^  =  0; 

.'.     c  =  o. 

H7  /-r3          73r\ 

^  (2) 


The  slope  is  greatest  when  x  =  o ; 


The  deflection  is  greatest  when  x  =  — ; 


48^7' 

4°.  In  the  following  table  7  denotes  the  moment  of  inertia 
of  the  largest  section : 


SPECIAL    CASES. 


305 


Uniform  Cross-Section. 

Greatest  Slope. 

Greatest 
Deflection. 

Fixed  at  one  end,  loaded  at  the  other 
Fixed  at  one  end,  loaded  uniformly     .     . 
Supported  at  ends,  load  at  middle  .     .     . 
Supported  at  ends,  uniformly  loaded  .     . 

i  Wl2 
2  El 
i  Wl2 
6^7 
i  Wl* 

i  Wl* 

~ZE! 

i  Wl* 
*EI 
i  Wl* 

'6  El 

i  Wl2 

&EI 
5  Wl* 

^EI 

384^7 

Uniform  Strength  and  Uniform   Depth, 
Rectangular  Section. 

Fixed  at  one  end,  load  at  the  other     .     . 

Fixed  at  one  end,  uniformly  loaded    .     . 

t 

Supported  at  both  ends,  load  at  middle   . 
Supported  at  both  ends,  uniformly  loaded, 

Wl2 
~EI 
i  Wl2 
2~EI 
i  Wl* 

i  Wl* 
*~EI 
i  Wl* 
4  El 
i  Wl* 

*$l 

i  Wl2 

32  El 

i  Wl* 

rt  El 

^El 

Uniform  Strength  and  Uniiorm  Breadth, 
Rectangular  Section. 

Fixed  at  one  end,  loaded  at  the  other, 
Supported  at  both  ends,  load  at  middle, 
Supported  at  both  ends,  uniformly  loaded 

wr 
2^i 
i  wr 

2    Wl* 

3  El 

i  wr 

4  El 

o  «^a 
0.098  -  — 

24  EI 

o  wr 
0.0:8  ~- 

3O6  APPLIED   MECHANICS. 

§197.    Deflection  with   Uniform    Bending-Moment If 

the  bending-moment  is  uniform,  then  M  is  constant ;  and,  if  7 
is  also  constant,  we  have 

.  _  M   C        _Mx 

but  when  x  =  -,  then  i  =  o; 

Ml 


2EI 


Ml        l\       dv 

•"•      l   —  ~WJ\  x  --  )  =  ~T 
EI\         2]       dx 


Mx2       lx 


the  constant  disappearing  because  v  =  o  when  x  —  o. 

Hence,  for  a  beam  where  the  bending-moment  is  uniform, 
we  have 

Ml         l  M   x*       lx 


and  for  greatest  slope  and  deflection,  we  have 

-Ml  Mil*       M  i  Ml2 


§  198.  Resilience  of  a  Beam.  —  The  resilience  of  a  beam 
is  the  mechanical  work  performed  in  deflecting  it  to  the  amount 
it  would  deflect  under  its  greatest  allowable  gradually  applied 
load.  In  the  case  of  a  concentrated  load,  if  W  is  the  greatest 
allowable  gradually  applied  load,  and  vl  the  corresponding 
deflection  at  the  point  of  application  of  the  load,  then  will  the 

W 
mean  value  of  the  load  that  produces  this  deflection  be  —  » 

W 
and  the  resilience  of  the  beam  will  be  ~vv 


SLOPE   AND   DEFLECTION  OF  A   BEAM. 


§  199.   Slope    and    Deflection    of   a    Beam  with   a   Con 
centrated    Load    not    at    the       <  > 

Middle.  —  Take,   as    the    next      °  A  _  _B 

case,  a  beam  (Fig.   228).     Let        <  a 

the  load  at  A  be  W,  and  dis- 

tance OA  —  a,  and  let  a  >  -. 

2  FIG.  228. 


W(l  -  a) 
<a     M  =  —  ^  -  '-  x, 


x>  a     M=  —  (/-  x), 


•••  x<a  '  = 


When  x  =.  o,     i  =  i0  =  undetermined  slope  at  O  ; 


and 


V/hen  x  =  o,     ^  =  o  ; 


W(l  -  a) 
61EI 


rr    \t,     &)  .  f      v 

•'•     V=^6iET2xZ  +  ^'  & 


To  determine  c,  observe  that  when  x  —  a,  this  value  of  i 
and  that  deduced  from  (i)  must  be  identical. 

Wa 


3°S  APPLIED   MECHANICS. 


Wa 


or 


and 


c2  -  -  -  lax)  +  4*  +  c. 
3 

To  determine  c,  observe  that  when  x  =  a,  this  value  of  z> 
and  (2)  must  be  identical ; 


/      a\       .  W(l  -  a) 

( )  +  io«  +  c  =  — V,/Z77  ;  «3  4-  ^ 

\       *  /  6/57 


2lEI\       3 


6/^7       6^7 

*3  -  3^  +'  l<?}  +  />.          (4) 
To  determine  z'0>  we  have  that  when  x  =  /,  v  =  o  ; 


Substituting  this  value  of  i0  in  the  equations  (i),  (2),  (3),  and 
(4),  we  obtain  for 


Wa 


SLOPE   AND  DEFLECTION  OF  A    BEAM.  309 


(3)  '•  = 


To  find  the  greatest  deflection,  differentiate  (2),  and  place 
the  first  differential  co-efficient  equal  to  zero  :  or,  which  is  the 
same  thing,  place  i  =  o  in  (i),  and  find  the  value  of  x;  then 
substitute  this  value  in  (2),  and  we  shall  have  the  greatest 
deflection. 
,  We  thus  obtain 


or 


3 
and  the  greatest  deflection  becomes 


=  _  Wa(l  -  a)  (2/  -  a)     2al  -  a2 

~~ 


§  2OO.  EXAMPLES. 

1.  In  example  i,  p.  294,  find  the  greatest  deflection  of  the  beam 
when  it  is  loaded  with  J  of  its  breaking-load,  assuming  E  =  1200000. 

2.  In  the  same  case,  find  what  load  will  cause  it  to  deflect  ^J^  of  its 
span. 

3.  What  will  be  the  stress  at  the  most  strained  fibre  when  this  occurs. 

4.  In  example  3,  p.  294,  find  the  load  the  beam  will  bear  without 
deflecting  more  than  ¥ fa  ot  its  span,  assuming  E  =  24000000. 

5.  Find  the  stress  at  the  most  strained  fibre  when  this  occurs. 

6.  In  example  6,  p.  295,  find  the  greatest  deflection  under  a  load 
J  the  breaking-load. 


3io 


APPLIED   MECHANICS. 


§201.   Deflection   and    Slope   under  Working-Load. If 

we  take  the  four  cases  of  deflection  given  in  the  first  part  of 
the  table  on  p.  305,  and  calling/  the  working  strength  of  the 
material,  and  y  the  distance  of  the  most  strained  fibre  from 
the  neutral  axis,  and  if  we  make  the  applied  load  the  working- 
load,  we  shall  have  respectively  — 


Wl=f- 

y 

M^fl 
2    '"  y 

iw_fi 

4  ~  y 

Wl      fi 

y 


fl 


iy 


4  -   -3-  =  - 


And  the  values  of  slope  and  deflection  will  become  respectively, 


Slope. 


Deflection. 


Slope. 


Deflection. 


2. 


From  these  values,  and  those  given  on  p.  305,  we  derive  the 
following  two  propositions  :  — 

i°.  If  we  have  a  series  of  beams  differing  only  in  length, 
and  we  apply  the  same  load  in  the  same  manner  to  each,  their 
greatest  slopes  will  vary  as  the  squares  of  their  lengths,  and 
their  greatest  deflections  as  the  cubes  of  their  lengths. 


SLOPE   AND   DEFLECTION  OF  RECTANGULAR  BEAMS.   311 


2°.  If,  however,  we  load  the  same  beams,  not  with  the  same 
load,  but  each  one  with  its  working-load,  as  determined  by 
allowing  a  given  greatest  fibre  stress,  then  will  their  greatest 
slopes  vary  as  the  lengths,  and  their  greatest  deflections  as  the 
squares  of  their  lengths. 

§  202.  Slope    and    Deflection   of   Rectangular   Beams 

bto  h 

If  the  beams  are  rectangular,  so  that  I  —  —  and  y  —  -,  the 

values  of  slope  and  deflection  above  referred  to  become  further 
simplified,  and  we  have  the  following  tables  :  — 


Given  Load  W. 

Working-Load. 
Greatest  Fibre  Stress  =/". 

Slope. 

Deflection. 

Slope. 

Deflection. 

1°. 

2°. 

3°- 
4°- 

6WP 

4073 

ft 

2/2 

Ebfc 

2\Vl* 

£tfr 

3  073 

Eh 

*  fl 

3  Eh 
i/2 

Ebte 
3  Wl2 

••J5&M 

I     073 

3£/i 
i  ft 
2  Eh 

*.£ 

*Eh 
i/2 
6  Eh 

Sfl* 

\Ebte 

i   Wl* 

I  Etfr 

5    073 

*  Ebte 

32  Ebh* 

ZEh 

^Eh 

So  that,  in  the  case  of  rectangular  beams  similarly  loaded  and 
-supported,  we  may  say  that  — 

Under  a  given  load  W,  the  slopes  vary  as  the  squares  of 
the  lengths,  and  inversely  as  the  breadths  and  the  cubes  of  the 
depths  ;  while  the  deflections  vary  as  the  cubes  of  the  lengths, 
and  inversely  as  the  breadths  and  the  cubes  of  the  depths. 


312  APPLIED   MECHANICS. 

On  the  other  hand,  under  their  working-loads,  the  slopes  vary 
directly  as  the  lengths,  and  inversely  as  the  depths  ;  while  the 
deflections  vary  as  the  squares  of  the  lengths,  and  inversely  as 
the  depths. 

§  203.   Beams  Fixed  at  the  Ends.  —  The  only  cases  which 
we  shall  discuss  here  are  the  two  following ;  viz.,  — 
i°.  Uniform  section  loaded  at  the  middle. 
2°.   Uniform  section,  load  uniformly  distributed. 
CASE  I.  —  Uniform   Section  loaded  at  the  Middle.  —  The 
fixing  at  the  ends  may  be  effected  by  building  the  beam  for 

some  distance  into  the  wall,  as 
shown  in  Fig.  229.  The  same 
result,  as  far  as  the  effect  on 
the  beam  is  concerned,  might 
be  effected  as  follows :  Hav- 

FIG.  229.  .  .  ,      . ,  , 

mg  merely  supported    it,   and 

placed  upon  it  the  loads  it  has  to  bear,  load  the  ends  outside 
of  the  supports  just  enough  to  make  the  tangents  at  the  sup- 
ports horizontal. 

These  loads  on  the  ends  would,  if  the  other  load  was  re- 
moved, cause  the  beam  to  be  convex  upwards :  and,  moreover, 
the  bending-moment  due  to  this  load  would  be  of  the  same 
amount  at  all  points  between  the  supports ;  i.e.,  a  uniform 
bending-moment.  Moreover,  since  the  effect  of  the  central 
load  and  the  loads  on  the  ends  is  to  make  the  tangents  over 
the  supports  horizontal,  it  follows  that  the  upward  slope  at  the 
support  due  to  the  uniform  bending-moment  above  described 
must  be  just  equal  in  amount  to  the  downward  slope  due  to  the 
load  at  the  middle,  which  occurs  when  the  beam  is  only  sup- 
ported. 

Hence  the  proper  method  of  proceeding  is  as  follows  :  — 
i°.  Calculate  the  slope  at  the  support  as  though  the  beam 
were  supported,  and  not  fixed,  at  the  ends ;  and  we  shall  have, 
if  we  represent  this  slope  by  iu  the  equation 


BEAMS  FIXED   AT  THE  ENDS. 


313 


Wl* 


2°.  Determine  the  uniform  bending-moment  which  would 
produce  this  slope. 

To  do  this,  we  have,  if  we  represent  this  uniform  bending- 
moment  by  Miy  that  the  slope  which  it  would  produce  would  be 

MJ 


and,  since  this  is  equal  to  *„  we  shall  have  the  equation 

MJ        Wl*  f  . 

•~^EJ-^'=Q  (3) 


.:    Jf,  =  -™  (4) 

This  is  the  actual  bending-moment  at  either  fixed  end  ;  and  the 
bending-moment  at  any  special  section  at  a  distance  x  from 
the  origin  will  be 


where  M  is  the  bending-moment  we  should  have  at  that  sec- 
tion if  the  beam  were  merely  supported,  and  not  fixed.  Hence, 
when  it  is  fixed  at  the  ends,  we  shall  have,  for  the  bending- 
moment  at  a  distance  x  from  O,  where  O  is  at  the  left-hand 
support, 

W        W 


When  x  =  -,  we  obtain,  as  bending-moment  at  the  middle, 

m 

M0  =  —  ;  (6) 

O 

and,  since  Ml  =  —Moy  it  follows  that  the  greatest  bending- 
moment  is 

Wl 


3T4  APPLIED   MECHANICS. 

this  being  the  magnitude  of  the  bending-moment  at  the  middle 
and  also  at  the  support. 

POINTS    OF    INFLECTION. 

The  value  of  M  becomes  zero  when 

x  =   -  and  when  x  —  —  : 
4  4' 

hence  it  follows  that  at  these  points  the  beam  is  not  bent,  and 
that  we  thus  have  two  points  of  inflection  half-way  between  the 
middle  and  the  supports. 

SLOPE    AND    DEFLECTION    UNDER    A    GIVEN    LOAD. 

We  shall  have,  as  before, 

Wx*       Wlx 


CM   , 
i  =  J  -  dx  = 


+  tf 


and  since,  when  x  =  o,  i  =  o, 
.*.    c  —  o 


.       dv         W  ,  .  N  i 

.*.       t   =   -   =   -  (2X2  —  lX}  (7) 

dx 

W 


the  constant  vanishing  because  v  =  o  when  x  =  o.     The  slope 

becomes  greatest  when  x  =  -,  and  the  deflection  when  x  =  -. 

4 
Hence  for  greatest  slope  and  deflection,  we  have 


64^7' 


(9) 


,     N 

v0  = — -.  (10) 

192^7 


BEAMS  FIXED   AT   THE   ENDS.  315 

SLOPE    AND    DEFLECTION    UNDER    THE    WORKING-LOAD. 

If  f  represent  the  working-strength  of  the  material  per 
square  inch,  and  if  W  represent  the  centre  working-load,  we 
shall  have 

W?  =SJ 
8    := 


•  ••'*<>--&,  <»>     -=-if 

CASE  II.  —  Uniform  Section,  Load  uniformly  Distributed.  — 
Pursuing  a  method  entirely  similar  to  that  adopted  in  the  former 
case,  we  have  — 

i°.  Slope  at  end,  on  the  supposition  of  supported  ends,  is 


2°.  Slope  at  end  under  uniform  bending-moment  MI  is 


Hence,  since  their  sum  equals  zero, 

Wl 
K  =  -->  (3) 

which  is  the  bending-moment  over  either  support. 
The  bending-moment  at  distance  x  from  one  end  is 

W  Wl 

*  f  W      /   -.  \  V  V    If  s       \ 

M=-(tx  --)--.  (4) 

This  is  greatest  when  x  =  o,  and  is  then .     Hence  great- 
est bending-moment  is,  in  magnitude, 

*?.  (5) 

12 


316  APPLIED   MECHANICS. 


POINTS    OF    INFLECTION. 

M  becomes  zero  when  x  —  -  ±  —  =.  (6) 

2  2^3 

Hence  the  two  points  of  inflection  are  situated  at  a  distance 
on  either  side  of  the  middle. 


SLOPE    AND    DEFLECTION. 


the  constant  vanishing  because  i  =  o  when  x  =  o. 


the  constant  vanishing  because  v  =  o  when  .#  =  o.     Hence  for 
greatest  slope  and  deflection  we  have,  i  is  greatest  when  x  = 

-(  i  ±  -7=\  and  v  is  greatest  when  x  =  -  ; 
2\          V  2 


SLOPE    AND    DEFLECTION    UNDER    WORKING-LOAD. 

For  working-load  we  have 


12 


(13) 


«-.  =  - 


B ENDING-MOMENT  AND  SHEARING-FORCE.  317 


EXAMPLES. 

1.  Given  a  4-inch  by  1 2-inch  yellow-pine  beam,  span  20  feet,  fixed 
at  the  ends ;  find  its  safe  centre  load,  its  safe  uniformly  distributed  load, 
and  its  deflection  under  each  load.     Assume  a  modulus  of  rupture  5000 
Ibs.  per  square  inch,  and  factor  of  safety  4.     Modulus   of  elasticity, 
i 200000. 

2.  Find  the  depth  necessary  that  a  4-inch  wide  yellow- pine  beam,  20 
feet  span,  fixed  at  the  ends,  may  not  deflect  more  than  one  four-hun- 
dredth of  the  span  under  a  load  of  5000  Ibs.  centre  load. 

§  204.  Variation  of  Bending-Moment  -with  Shearing- 
Force.  —  If,  in  any  loaded  beam  whatever,  M  represent  the 
b ending-moment,  and  F  the  shearing-force  at  a  distance  x  from 
the  origin,  then  will 

F^^'  (l) 

Proof  (a).  —  In  the  case  of  a  cantilever  (Fig.  230),  assume 
the  origin  at  the  fixed  end  ;  then,  if  M 
represent  the  bending-moment  at  a 
distance  .#  from  the  origin,  and  M  -\-  AJ/ 
that  at  a  distance  x  -f-  &x  from  the 
origin,  we  shall  have  the  following 

FIG.  230. 

equations  :  — 

M=  -2        W(a-  x}, 

X  =  X 

x  =  l 

M+  AJ/=  —2,        W(a—  x  —  A#)  nearly. 

X  =  X 

a  being  the  co-ordinate  of  the  point  of  application  of  W, 

W  nearly 

=  X 

x  =  l 

S         W; 


3^8  APPLIED   MECHANICS. 

and,  if  we  pass  to  the  limit,  and  observe  that 

x  =  l 


I        W, 

X  =  X 


we  shall  obtain 


(b)  In  the  case  of  a  beam  supported  at  the  ends  (Fig.  231), 
4_A__>  assume  the  origin   at  the  left-hand 
^4       ""7s*   end,  and  let  the  left-hand  support- 
ing-force be  S ;  then,  if  a  represent 
FlG-  23i-  the  distance  from  the  origin  to  the 
point  of  application  of    W,  we  shall  have  the  equations 


=  Sx  -  2       W(x  -  a), 


M  +  bM  =  S(x  +  A*)  -  2        W(x  -  a  +  &x)  nearly. 

x  —  o 

Hence,  by  subtraction, 

X  =  X 

=  S  .  A*  —  2       W&x  nearly 


X  =  O 


=  S  -  2        W  nearly; 

and,  if  we  pass  to  the  limit,  and  observe  that 
F  =  S  -  £    *  W, 

X  =  O 

we  shall  obtain 

dM "_  F 

as  before. 


LONGITUDINAL   SHEARING    OF  BEAMS.  319 

§  205.  Longitudinal  Shearing  of  Beams. — The  resistance 
of  a  beam  to  longitudinal  shearing  sometimes  becomes  a  mat- 
ter of  importance,  especially  in  timber,  where  the  resistance  to 
shearing  along  the  grain  is  very  small.  We  will  therefore  pro- 
ceed to  ascertain  how  to  compute  the  intensity  of  the  longi- 
tudinal shear  at  any  point  of  the  beam,  under  any  given  load ; 
as  this  should  not  be  allowed  to  exceed  a  certain  safe  limit,  to 

be  determined  experimentally.     Assume  a  

section  ^^(Fig.  232)  at  a  distance  x  from 
the  origin,  and  let  the  bending-moment  at 
that  section  be  M.  Let  the  section  BD  be 

at  a  distance  x  +  A^r  from  the  origin,  and  — 

o  o 

let  the  bending-moment  at  that  section  be  FIG.  232. 


b  / 


7 


M  -f- 

Let  y0  be  the  distance  of  the  outside  fibre  from  the  neutral 
axis  ;  and  let  ca  =  y^  be  the  distance  of  a,  the  point  at  which 
the  shearing-force  is  required,  from  the  neutral  axis. 

Consider  the  forces  acting  on  the  portion  ABba,  and  we 
J^all  have  — 

'My0 
i°.   Intensity  of  direct  stress  at  A  =  -=^. 

2°.   Intensity  of  direct  stress  at  a  unit's  distance  from  neu- 

,      •         M 
tral  axis  -    —jr. 

My 
3°.   Intensity  of  direct  stress  at  e,  where  ce  —  y,  is  —=-. 

(M  -f-  &M}y 
So,  likewise,  intensity  of  direct  stress  at  f  is  -      —  j  — 

Therefore,  if  z  represent  the  width  of  the  beam  at  the  point 
e,  we  shall  have  — 

M  (*y* 

Total  stress  on  face  Aa  —  -=•  /    yzdy, 

1  J 


M 
Total  stress  on  face  Bb  ='  -  j  -  I    yzdy  ; 


320  APPLIED   MECHANICS. 


T-vrr  ^M  Cy° 

.'.     Difference  =   -  I    yzdy  : 
I  Jyi 

and  this  is  the  total  horizontal  force  tending  to  slide  the  piece 
AabB  on  the  face  ab. 

Area  of  face  ab,  if  zt  is  its  width,  is 


therefore  intensity  of  shear  at  a  is  approximately 

° 

yzdy 


or  exactly  (by  passing  to  the  limit) 

dM\ 

&)  C*> 

j-l    yzdy. 

M      */J| 

dM 
And,  observing  that  F  =  -j-,  this  intensity  reduces  to 


(I) 

We  may  reduce  this  expression  to  another  form  by  observ- 
ing, that,  if  yz  represent  the  distance  from  c  to  the  centre  of 
gravity  of  area  Aa,  and  A  represent  its  area,  we-  have 

f*v 

I    yzdy  =  y2A  ; 
t/jj/j 

therefore  intensity  of  shear  (at  distance  yl  from  neutral  axis)  at 
point  a  = 


This  may  be  expressed  as  follows  :  — 


LONGITUDINAL   SHEARING    OF  BEAMS.  321 

Divide  the  shearing-force  at  the  section  of  the  beam  under 
consideration,  by  the  product  of  the  moment  of  inertia  of  the 
section  and  its  width  at  the  point  where  the  intensity  of  the 
shearing-force  is  desired,  and  multiply  the  quotient  by  the  statical 
moment  of  the  portion  of  the  cross-section  between  the  point  in 
question  and  the  outer  fibre  ;  this  moment  being  taken  about  the 
^neutral  axis.  The  result  is  the  required  intensity  of  shear. 

The  last  factor  is  evidently  greatest  at  the  neutral  axis  ; 
hence  the  intensity  of  the  shearing-force  is  greatest  at  the 
neutral  axis. 

LONGITUDINAL   SHEARING  OF   RECTANGULAR  BEAMS. 

Eor  rectangular  beams,  we  have 

bfc 
/=—  ,  ,,  =  *. 

Hence  formula  (2)  becomes 


For  the  intensity  at  the  neutral  axis,  we  shall  have,  therefore, 
1  2.F  //i  bh\      .  3  F 

~frfc  \4  ~2/    ==    2  Ttt  k) 

since  for  the  neutral  axis  we  have 


h  bh 

y2=-     and     A  =  -. 


EXAMPLES. 


i.  What  is  the  intensity  of  the  tendency  to  shear  at  the  neutral  axis 
of  a  rectangular  4-inch  by  12  -inch  beam,  of  14  feet  span,  loaded  at  the 
middle  with  5000  Ibs. 


322 


APPLIED   MECHANICS. 


2.  What  is  that  of  the  same  beam  at  the  neutral  axis  of  the  cross- 
section  at  the  support,  when  the  beam  has  a  uniformly  distributed  load 
of  12000  Ibs. 

3.  What  is  that  of  a  Q-inch  by  1 4-inch  beam,  20  feet  span,  loaded 
with  15000  Ibs.  at  the  middle. 


§  206.  Strength  of  Hooks.  —  The  following  is  the  method 
to  be  pursued  in  determining  the  stresses  in  a 
hook  due  to  a  given  load  ;  or,  vice  versa,  the 
proper  dimensions  to  use  for  a  given  load. 

Suppose  (Fig.  233)  a  load  hung  -at  E  ;  the 
load  being  P,  and  the  distances 
AB  —  n\ 


O  being  the  centre  of  -gravity  of  this 
section,  conceive  two  equal  and  opposite 
forces,  each  equal  and  parallel  to  P,  acting 
at  <9. 

Let  A  =  area  of  section,  and  let  /  =  its 
moment  of  inertia  about  CD  (BCDF  represents  the  section 
revolved  into  the  plane  of  the  paper);  then  — 

i°.  The  downward  force  at  O  causes  a  uniformly  distributed 
stress  over  the  section,  whose  intensity  is 

P 

A  =  ^- 

2°.  The  downward  force  at  E  and  the  upward  force  at  O 
constitute  a  couple,  whose  moment  is 


and  this  is  resisted,  just  as  the  bending-moment  in  a  beam,  by 
a  uniformly  varying  stress,  producing  tension  on  the  left,  and 
compression  on  the  right,  of  CD. 


SHORT  STRUTS. 


323 


If  we  call  /a  the  greatest  intensity  of  the  tension  due  to 
this  bending-moment,  viz.,  that  at  B,  we  have 


/ 

and  if  /3  denote  the  greatest  intensity  of  the  compression  due 
to  the  bending  moment,  viz.,  that  at  F,  we  have 


J 

therefore  the  actual  greatest  intensity  of  the  tension  is 

*i  J. 

and  this  must  be  kept  within  the  working  strength  if  the  load 
is  to  be  a  safe  one ;  and  so  also  the  actual  greatest  intensity 
of  the  .compression,  viz.,  that  at  /%  is,  when/,  >  A» 

A  =  A  -A  =         ^     ^  -  ^, 
which  must  be  kept  within  the  working  strength  for  compression. 

§  207.  Short  Struts The  case  of  a  short  strut,  with  the 

load  applied  at  some  point  other  than  the 
centre  of  gravity  of  the  section,  is  similar  to 
that  of  the  hook.  Thus,  let  O'  (Fig.  234)  be 
the  centre  of  gravity  of  the  lower  section, 
and  let  A' 0'  —  xv 

Conceive  two  equal   and  opposite  forces 
at  O'j  each  equal  and  parallel  to  P,  and  we 

FIG.  234. 

have  — 

i°.  Downward  force  along  line  OO'  causes  uniform  stress 
of  intensity, 


B 

!A      O 

i 

B1 

jA1      O1 

2°.  The  other  two  form  a  couple,  whose  moment  is 


324  APPLIED   MECHANICS. 

therefore  the  greatest  intensity  of  the  compression  due  to  this 
will  be  that  at  By  or 


where  a  =  O'B  '  .     Hence  total  greatest  intensity  is 


or,  if  we  write 

/  =  Af, 

where  p  =  radius  of  gyration  of  lower  section  about  the  axis 
through  O'  perpendicular  to  the  plane  of  the  paper,  we  have 


and   this   should   be   kept  within   the  limits  of   the  working- 
strength  of  the  material. 

EXAMPLES. 

1.  Given  a  cylindrical  column  of  8  inches  diameter,  and  let  x0  =  2 
inches  :  find  greatest  stress  per  square  inch  under  a  load  of  100000  Ibs. 

2.  Given  P  =   200000  Ibs.,  x0  =  2  inches  :    find  diameter  of  a 
yellow-pine    strut    suitable    to    bear   the   load,  with  factor  of  safety  4. 
Compressive  strength  of  yellow  pine  =  4400  Ibs.  per  square  inch. 

§  208.  Strength  of  Columns. — The  formulae  most  com- 
monly used  for  the  strength  of  columns  are  of  three  kinds  ;  viz., 

i°.  Euler's  formulae,  where  it  is  assumed,  that,  for  any  given 
material,  there  is  a  certain  definite  ratio  of  length  to  diameter, 
below  which  a  column  will, give  way  by  direct  crushing,  while 
one  whose  ratio  of  length  to  diameter  is  greater  will  give  way 
wholly  by  transverse  bending. 

2°.  Hodgkinson's  empirical  formulae,  based  upon  his  experi- 
ments upon  small  columns  of  a  variety  of  ratios  of  length  to 
diameter. 


GORDON'S  FORMULA   FOR   COLUMNS.  $2$ 

3°.  Gordon's  formulae,  where  it  is  assumed  that  all  columns 
give  way  by  a  combination  of  crushing  and  bending. 

It  is  very  much  to  be  regretted  that  none  of  these  sets  of 
formulae  are  borne  out  by  experiment  upon  the  large  scale, 
and  that  thus  far  we  have  no  formulae  for  columns  that  are 
borne  out  generally.  Euler's  are  evidently  faulty  in  the  funda- 
mental assumption  ;  Hodgkinson's  experiments  were  made  on 
small  columns,  and  do  not  agree  well  with  those  on  large  ones  ; 
Gordon's,  or,  as  they  are  otherwise  called,  Rankine's,  are  prob- 
ably correct  in  their  fundamental  assumption,  but  there  is  a 
serious  lapse  in  the  reasoning  by  which  they  are  deduced. 

The  formulae  most  frequently  used  in  American  practice 
are  those  of  Gordon.  Hence  we  will  take  those  first. 

§  209.    Gordon's    Fqrmulse    for    Columns,      (a)    Column 
fixed  in  Direction  at  Both  Ends.  —  Let  CAD  be  the  cen- 
tral axis  of  the  column,  P  the  breaking-load,  and  v  the        "F 
greatest  deflection,  AB.     Conceive  at  A  two  equal  and        /[ 
opposite  forces,  each  equal  to  P  ;  then  — 

i°.  The  downward  force  at  A  causes  a  uniformly  dis- 
tributed stress  over  the  section,  of  intensity, 


2°.  The  downward  force  at  C  and  the  upward  force  , 

.  .  ,  rlG.  235. 

at  A  constitute  a  couple,  whose  moment  is 

M  —  Pv  ; 

and  this  is  resisted,  just  as  the  bending-moment  in  a  beam,  by 
a  uniformly  varying  stress,  producing  compression  on  the  right, 
and  tension  on  the  left,  of  A. 

If  we  call  p2  the  greatest  intensity  of  the  compression  due 
to  this  bending,  we  have 


326  APPLIED   MECHANICS. 

where  y  =  distance  from  the  neutral  axis  to  the  most  strained 
fibre  of  the  section  at  A.  Then  will  the  greatest  intensity  of 
the  stress  of  compression  at  section  A  be 


and,  since  P  is  the  breaking-load,  /  must  be  equal  to  the  break- 
ing-strength for  compression  per  square  inch  =f. 
Hence 


where  p  =  smallest  radius  of  gyration  of  section  at  A. 

Thus  far  the  reasoning  appears  sound  ;  but  in  the  next  step 
it  is  assumed,  that  because,  in  a  loaded  beam,  the  greatest 
deflection  under  the  breaking-load  varies  as  the  square  of  the 
length,  and  inversely  as  the  distance  from  the  neutral  axis  to 
the  most  strained  fibre,  therefore  in  this  case  it  is  assumed 

that  we  must  have  also 

/2 


or 


v  =  --  , 
c  y' 


where  c  is  a  constant  to  be  determined  by  experiment.     Hence 


therefore,  substituting  this  in  (i), 

' 


(.) 


GORDON'S  FORMULAE   FOR    COLUMNS. 


327 


which  is  the  required  formula  for  a  column  fixed  in  direction 
at  both  ends. 

(b)  Column  hinged  at  the  Ends.  —  It  is  assumed  in  the 
previous  case  that  the  points  of  inflection  are  halfway 
between  the  middle  and  the  ends,  and  hence  that,  by 
taking  the  middle  half,  we  have  the  case  of  bending  of  a 
column  hinged  at  the  ends  (Fig.  236).  Hence,  to  obtain 
the  formula  suitable  for  this  case,  substitute,  in  (2),  2/ 
for  /,  and  we  obtain  FIG.  236. 


(3) 


(c)  Column  fixed  at  One  End  and  hinged  at  the  Other  (Fig. 
237).  —  In  this  case  we  should,  in  accordance  with  these 
assumptions,  take  f  of  the  column  fixed  in  direction  at 
both  ends ;  hence,  to  obtain  the  formula  for  this  case, 
substitute,  in  (2),  \l  for  /,  and  we  thus  obtain 


FIG.  237. 


i6/2' 
9'P2 


(4) 


Rankine  gives,  for  values  of  f  and  c,  the  following,  based 
upon  Hodgkinson's  experiments  :  — 


/ 
(Ibs.  per  sq.  in.). 

c. 

36000 

36000 

Cast-iron                             

80000 

6400 

72OO 

3000 

328  APPLIED   MECHANICS. 

§  210.   Euler's  Rules  for  the  Strength  of  Columns. — The 

following  are  the  rules  for  determining  the  strength  of  a  col- 
umn of  uniform  cross-section,  according  to  Euler :  — 

(a)  Column  fixed  in  Direction  at  One  End  only,  which  bends, 
as  shown  in  the  Figure. 

i°.  Calculate  the  breaking-load  on  the  assumption  that  the 
column  will  give  way  by  direct  compression.  This  will  be 

P*=fA,  (i) 

where  f  =  crushing-strength  per  square  inch,  and  A  =  area 
of  cross-section  in  square  inches. 

2°.  Calculate  the  load  that  would  break  the  column  if  it 
were  to  give  way  by  bending,  by  means  of  the  following  for- 
mula: — 


ft*  ft 


where  E  =  modulus  of  elasticity  of  the  material,  7  =  smallest 
moment  of  inertia  of  the  cross-section,  and  /  =  length  of 
column. 

Then  will  the  actual  breaking-strength,  according  to  Euler, 
be  the  smaller  of  these  two  results. 

To  deduce  the    latter  formula,  assume    the   origin    at   the 
hinged  end,  and  take  x  vertical  and  y  horizontal. 

Let  p  =  radius  of  curvature  at  point  (x,  y),  and 
let  M  —  bending-moment  at  the  same  point. 

Then  we  shall  have,  just  as  was  shown   in    the 
case  of  the  deflection  of  beams, 


But  as  was  there  shown, 


M       Py 


EULER'S  RULES  FOR  STRENGTH  OF  COLUMNS. 


/dy    d*y,         P  f  dy 
i*Ji**W 


+ 


and,  since  for 


dy 

:/ 

dx 


(4) 


And  since,  when 

x  =  o, 
and  we  have 


=  o, 


c- 


.-.    c  =  o, 


(5) 


When  y  =  a,  we  know  that  ;tr  =  //  hence,  substituting  in 
,  we  have 


or 


p=  - 


(6) 


33O  APPLIED   MECHANICS. 

(b)   Column  hinged  at  Both  Ends  (Fig.  236). 

i°.  Calculate  the  crushing-load,  as  before,  from  the  formula 

P,  =fA. 

2°.  Calculate  the  load  that  would  break  it,  if  it  were  to  give 
way  wholly  by  transverse  bending,  from  the  formula 


this  being  derived  from  (2)  or  (6)  by  substituting  -  for  //  the 

reasoning  being  the  same  for  this  substitution  as  was  adopted 
with  Gordon's  formula. 

(c)  Column  fixed  in  Direction  at  Both  Ends  (Fig.  235).  — 
We  have  for  the  crushing-load  the  same  formula  as  before  ; 
viz.,  — 


and  for  the  bending  we  have 

P2  =  /,  (8) 


this  being  obtained  from  (2)  or  (6)  by  substituting  -  for  /. 

(d}  These  rules  may  be  summed  up  as  follows  :  — 
i°.  Calculate  the  crushing-load  by  the  formula 


2°.  Calculate   the   load   that  would   break  the   column  by 
bending,  from  the  following  formulae  :  — 


(•>  P.  =  M 


HODGKINSON'S  RULES  FOR  STRENGTH  OF  COLUMNS.    331 

if  fixed  in  direction  at  one  end  only  ; 
OS)  />,  =  (^EI 

if  hinged  or  rounded  at  both  ends  ; 


if  fixed  in  direction  at  both  ends. 

Then  will  the  actual  breaking-strength  be  the  least  of  the 
two  results. 

(e)  In  order  to  ascertain  the  length  where  incipient  flexure 
occurs,  according  to  this  theory  we  should  place  the  two  results 
equal  to  each  other,  and  from  the  resulting  equation  determine  /. 
We  should  thus  obtain,  for  the  three  cases  respectively,  — 

,    ^  ,  .       El 

(a) 


08) 

(r) 


Hence  all  columns  whose  length  is  less  than  that  given  in 
these  formulae  will,  according  to  Euler,  give  way  by  direct 
crushing  ;  and  those  of  greater  length,  by  bending  only. 

§  211.  Hodgkinson's  Rules  for  the  Strength  of  Columns. 
—  Eaton  Hodgkinson  made  a  large  number  of  tests  of  small 
columns,  especially  of  cast-iron,  and  deduced  from  these  tests 
certain  empirical  formulae.  These  tests  form,  even  at  the  pres- 
ent time,  the  basis  of  the  most  used  formulae  for  the  strength 
of  columns.  The  strength  of  pillars  of  the  ordinary  sizes  used 
in  practice  has  been  computed  by  means  of  Hodgkinson's  for- 


332  APPLIED   MECHANICS. 

mulae,  and  tabulated  by  Mr.  James  B.  Francis  :  and  we  find  in 
his  book  the  following  rules  for  the  strength  of  solid  cylindrical 
pillars  of  cast-iron,  with  the  ends  flat;  i.e.,  "finished  in  planes 
perpendicular  to  the  axis,  the  weight  being  uniformly  distrib- 
uted on  these  planes." 

For  pillars  whose  length  exceeds  thirty  times  their  diameter, 
he  gives  the  formula, 

,  (i) 


where  D  =  diameter  of  column  in  inches,  /  =  length  in  feet, 
W  =  breaking-weight  in  pounds. 

If,  on  the  other  hand,  the  length  does  not  exceed  thirty  times 
the  diameter,  he  gives,  for  the  breaking-weight,  the  following 
formula  :  — 

W  -       Wc 


where  W  =  breaking-weight  that  would  be  derived  from  the 
preceding  formula,  W  =.  actual  breaking-weight,  c  =  weight 
which  would  crush  the  pillar,  or 

/  T-V_\ 

(3) 

For  hollow  cast-iron  pillars,  if  D  =  external  diameter  in  inches, 
d  =  internal  diameter  in  inches,  we  should  have,  in  place  of  (i), 


•txr  o  —  /    \ 

W  =  99318 _ ,  (4) 

and  in  place  of  (3), 

c  =  109801  ^-^2).  (S) 

For  very  long  wrought-iron  pillars,  Hodgkinson  found  the 
strength  to  be  1.745  times  that  of  a  cast-iro*n  pillar  of  the  same 
dimensions  ;  but,  for  very  short  pillars,  he  found  the  strength  of 


_G K 


H      I  P 

LJD 

FIG.  230. 


STRENGTH  OF  SHAFTING.  333 

the  wrought-iron  pillar  very  much  less  than  that  of  the  cast-iron 
one  of  the  same  dimensions.  With  a  length  of  30  diameters 
and  flat  ends,  the  wrought-iron  exceeded  the  cast-iron  by  about 
ten  per  cent. 

§  212.  Strength  of  Shafting.  —  The  usual  criterion  for  the 
strength  of  shafting  is,  that  it  shall  be  sufficiently  strong  to 
resist  the  twisting  to  which  it  is  exposed  in  the  transmission  of 
power. 

Proceeding  in  this  way,  let  EF  (Fig.  239)  be  a  shaft,  AB  the 
driving,  and  CD  the  following,  pulley. 
Then,  if  two  cross-sections  be  taken 
between  these  two  pulleys,  the  por- 
tion of  the  shaft  between  these  two 
cross-sections  will,  during  the  trans- 
mission of  power,  be  in  a  twisted  con- 
dition ;  and  if,  when  the  shaft  is  at 

rest,  a  pair  of  vertical  parallel  diameters  be  drawn  in  these  sec- 
tions, they  will,  after  it  is  set  in  motion,  no  longer  be  parallel, 
but  will  be  inclined  to  each  other  at  an  angle  depending  upon 
the  power  applied.  Let  GH  be  a  section  at  a  distance  x  from 
O,  and  let  KI  be  another  section  at  a  distance  x  -f-  dx  from  O. 
Then,  if  di  represent  the  angle  at  which  the  originally  parallel 
diameters  of  these  sections  diverge  from  each  other,  and  if  r  = 
the  radius  of  the  shaft,  we  shall  have,  for  the  length  of  an  arc 
passed  over  by  a  point  on  the  outside, 

rdi  ; 

and  for  the  length  of  an  arc  that  would  fbe  passed  over  if  the 
sections  were  a  unit's  distance  apart,  instead  of  dx  apart, 

rdi  _      di 

dx         dx      » 

This  is  called  the  strain  of  the  outer  fibres  of  the  shaft,  as  it 
is  the  distortion  per  unit  of  length  of  the  shaft. 


334  APPLIED   MECHANICS. 

In  all  cases  where  the  shaft  is  homogeneous  and  symmet- 
rical,, if  i  is  the  angle  of  divergence  of  two  originally  parallel 
diameters  whose  distance  apart  is  _r,  we  shall  have  the  strain, 

di         i 

v  =  r—  =  r-. 

ax        x 

This  also  is  the  tangent  of  the  angle  of  the  helix. 

A  fibre  whose  distance  from  the  axis  of  the  shaft  is  unity, 
will  have,  for  its  strain, 

f*L     •  L 

dx        x 

A  fibre  whose  distance  from  the  axis  of  the  shaft  is  p,  will  have, 
for  its  strain, 

di         i 

v  —  p—  =  p-. 

dx         x 

Fixing,  now,  our  attention  upon  one  cross-section,  GH,  we  have 
that  the  strain  of  a  fibre  at  a  distance  p  from  the  axis  (p  varying, 
and  being  the  radius  of  any  point  whatever)  is 


where  -  is  a  constant  for  all  points  of  this  cross-section. 

Hence,  assuming  Hooke's  law,  "  Ut  tensio  sic  vis"  we  shall 
have,  if  C  represent  the  shearing  modulus  of  elasticity,  that  the 
stress  of  a  fibre  whose  distance  from  the  axis  is  p,  is 


which  quantity  is  proportional  to  p,  or  varies  uniformly  from  the 
centre  of  the  shaft. 

The  intensity  at  a  unit's  distance  from  the  axis  is 


to- 


STRENGTH   OF  SHAFTING.  335 

and  if  we  represent  this  by  a,  we  shall  have  for  that  at  a  dis- 
tance p  from  the  axis, 

P  =  <*P- 

Hence  we  shall  have  (Fig,  240),  that,  on  a  small 
area,  V  / 

dA  =  dp(pdO)  =  pdpdO,  i^. 

the  stress  will  be 

pdA  =  apdA  —  ap2dpd0. 

The  moment  of  this  stress  about  the  axis  of  the  shaft  i§ 

ppdA  =  ap2dA  =  ap*dpd6, 

and  the  entire  moment  of  the  stress  at  a  cross-section  is 
afpdA  =  affptdpdB  =  al, 


where  7  =  fp2dA  is  the  moment  of  inertia  of  the  section  about 
the  axis  of  the  shaft. 

This  moment  of  the  stress  is  evidently  caused  by,  and  hence 
must  be  balanced  by,  the  twisting-moment  due  to  the  pull  of  the 
belt.  Hence,  if  M  represent  the  greatest  allowable  twisting- 
moment,  and  a  the  greatest  allowable  intensity  of  the  stress  at 
a  unit's  distance  from  the  axis,  we  shall  have 

M  =  al  =  -  /. 
P 

If  /  is  the  safe  working  shearing-strength  of  the  material 
per  square  inch,  we  shall  have  f  as  the  greatest  safe  stress  per 
square  inch  at  the  outside  fibre,  and  hence 


will  be  the  greatest  allowable  twisting-moment. 


3 3^  APPLIED   MECHANICS. 


For  a  circle,  radius  ry 


For  a  hollow  circle,  outside  radius  rn  inside  radius  r2, 


Moreover,  if  the  dimensions  of  a  shaft  are  given,  and  the 
actual  twisting-moment  to  which  it  is  subjected,  the  stress  at  a 
fibre  at  a  distance  p  from  the  axis  will  be  found  by  means  of  the 

formula 

MP 


The  more  usual  data  are  the  horse-power  transmitted  and 
the  speed,  rather  than  the  twisting-moment 

If  we  let  P  =  force  applied  in  pounds  and  R  =  its  leverage 
in  inches,  as,  for  instance,  when  P=  difference  of  tensions  of 
belt,  and  R  =  radius  of  pulley,  we  have 

M=  P  .R; 

and    if    HP  —   number    of    horses-power    transmitted,     and 
N  =  number  of   turns  per   minute,  then 


12    X    3300O 


EXAMPLE. 


Given  working-strength  for  shearing  of  wrought-iron  as  10000 
Ibs.  per  square  inch  ;  find  proper  diameter  of  shaft  to  transmit 
2O-horse  power,  making  100  turns  per  minute. 


TRANSVERSE  DEFLECTION  OF  SHAFT.  337 

Angle  of  Torsion.—  From  the  formula,  page  336,  p  —  "-j  , 


combined  with 


we  have 


p  —  ap  =  Cp-, 

oc 


MX 


which  gives  the  circular  measure  of  the  angle  of  divergence  of 
two  originally  parallel  diameters  whose  distance  apart  is  x  ;  the 
twisting-moment  being  M,  and  the  modulus  of  shearing  elas- 
ticity of  the  material,  C. 

EXAMPLES. 

1.  Find  the  angle  of  twist  of  the  shaft  given  in  example  i,  §  212, 
when  the  length  is  10  feet,  and  C  =  8500000. 

2.  What  must  be  the  diameter  of  a  shaft  to  carry  80  horses-power, 
with  a  speed  of  300  revolutions  per  minute,  and  factor  of  safety  6,  break- 
ing shearing-strength  of  the  iron  per  square  inch  being  50000  Ibs. 

§  213.  Transverse  Deflection  of  Shafts.  —  In  determining 
the  proper  diameter  of  shaft  to  be  used  in  any  given  case,  we 
ought  not  merely  to  consider  the  resistance  to  twisting,  but 
also  the  deflection  under  the  transverse  load  of  the  belt-pulls, 
weights  of  pulleys,  etc.  This  deflection  should  not  be  allowed 
to  exceed  -j^-  of  an  inch  per  foot  of  length.  Hence  the  de- 
flection should  be  determined  in  each  case. 

The  formulae  for  computing  this  deflection  will  not  be  given 
here,  as  the  methods  to  be  pursued  are  just  the  same  as  in  the 
case  of  a  beam,  and  can  be  obtained  from  the  discussions  on 
that  subject. 


33$  APPLIED   MECHANICS. 

§  214.  Combined  Twisting  and  Bending.  —  The  most  com- 
mon case  of  a  shaft  is  for  it  to  be  subjected  to  combined  twisting 
and  bending.  The  discussion  of  this  case  involves  the  theory 
of  elasticity,  and  will  not  be  treated  here ;  but  the  formulae  com- 
monly given  will  be  stated,  without  attempt  to  prove  them  until 
a  later  period.  These  formulae  are  as  follows  :  — 
Let  Ml  =  greatest  bending-moment, 

M2  =  greatest  twisting-moment, 

r     =  external  radius  of  shaft, 

/     =  moment  of  inertia  of  section  about  a  diameter, 

?rr4 

for  a  solid  shaft  /  = , 

4 

f     =  working-strength  of  the  material  =  greatest  al- 
lowable stress  at  outside  fibre ; 
then 

i°.  According  to  Grashof, 


/         ji  8M*  +    §V       i2    "I"         2    j- 

2°.  According  to  Rankine, 

§  215.  Springs. — The  object  of  this  discussion  is  to  enable  us 
to  answer  the  following  three  questions :  (a)  Given  a  spring, 
to  determine  the  load  that  it  can  bear  without  producing  in  the 
metal  a  maximum  fibre  stress  greater  than  a  given  amount. 
(&)  Given  a  spring,  to  determine  its  displacement  (elongation, 
compression,  or  deflection)  under  any  given  load,  (c)  Given  a 
load  P  and  a  displacement  dl  ;  a  spring  is  to  be  made  of  a 
given  material  such  that  the  load  P  shall  produce  the  displace- 
ment tf, ,  and  that  the  metal  shall  not,  in  that  case,  be  subjected 
to  more  than  a  given  maximum  fibre  stress.  Determine  the 
proper  dimensions  of  the  spring. 


SPRINGS.  339 


There  are  practically  only  two  cases  to  be  considered  as  far 
as  the  manner  of  resisting  the  load  is  concerned.  In  the  first, 
the  spring  is  subjected  to  transverse  stress,  and  is  to  be  calcu- 
lated by  the  ordinary  rules  for  beams.  In  the  second,  the 
spring  is  subjected  to  torsion,  and  the  ordinary  rules  for  re- 
sistance to  torsion  apply.  It  is  true  that  in  most  cases  where 
the  spring  is  subjected  to  torsion  there  is  also  a  small  amount 
of  transverse  stress  in  addition  to  the  torsion  ;  but  in  a  well- 
made  spring  this  transverse  stress  is  of  very  small  amount,  and 
we  may  neglect  it  without  much  error. 

We  will  begin  with  those  cases  where  the  spring  is  subjected 
to  torsion,  and  for  all  cases  we  shall  adopt  the  following  nota 
tion  : 

P  =  load  on  spring  producing  maximum  fibre  stress/; 

f  =  greatest  allowable  maximum  fibre  stress  for  shearing ; 

C  =  shearing  modulus  of  elasticity ; 

x  =  length  of  wire  forming  the  spring ; 
Ml  =  greatest  twisting  moment  under  load  P] 

L  =  any  load  less  than  the  limit  of  elasticity ; 

M  =  twisting  moment  under  this  load  ; 

/  =  maximum  fibre  stress  under  load  L ; 

p  =  distance  from  axis  of  wire  to  most  strained  fibre  ; 

/  =  moment  of  inertia  of  section  about  axis  of  wire ; 

it  =  angle  of  twist  of  wire  under  load  P ; 
i  =  angle  of  twist  of  wire  under  load  L ; 

V  =  volume  of  spring  ; 

£T  =  displacement  of  point  where  load  is  applied  when  load 
isP; 

$  =  displacement  of  point  where  load  is  applied  when  load 
isZ. 

Then  from  pages  335  and  337  we  obtain  the  following  four 
formulae : 


34°  APPLIED  MECHANICS. 


._  MX 

*~  ~cT' 


*:  =  £'.  (3) 


'--Z7-  <4) 

These  four  formulae  will  enable  us  to  solve  all  the  cases  of 
springs  subjected  to  torsion  only.  Moreover,  in  the  cases 
which  we  shall  discuss  under  this  head,  the  wire  will  have  either 
a  circular  or  a  rectangular  section :  in  the  former  case  we  will 
denote  its  diameter  by  d,  and  we  shall  then  have 

T      rtd*  d 

/  = and          p  =  — ; 

32  2 

while  in  the  latter  case  we  will  denote  the  two  dimensions  of 
the  rectangle  by  b  and  h,  respectively,  and  we  shall  then  have 


We  will  now  proceed  to  determine  the  values  of  P,  #,  6l ,  and 
V  in  each  of  the  following  four  cases,  all  of  which  are  cases  of 
torsion : 

CASE  I.  Simple  round  torsion  wire. — rLet  AB,  the  leverage 
of  the  load  about  the  axis,  be  R ;  then  we  shall  have 

M  =  LR,        M,  =  PR  ; 
and  we  readily  obtain  from  the  formulas  (i),  (2),  (3),  and  (4) 


SPRINGS.  341 


and  from  these  we  readily  obtain 


C         4  C 

(8) 


CASE  2.  Simple  rectangular  torsion  wire.  —  In  this  case  we 
readily  obtain 


iRx       f 


n 


(12) 

CASES  3  and  4.  Helical  springs  made  of  round  and  of  rec- 
tangular wire  respectively. — A  helical  spring  may  be  used  either 
in  tension  or  in  compression.  In  either  case  it  is  important 
that  the  ends  should  be  so  guided  that  the  pair  of  equal  and 
opposite  forces  acting  at  the  ends  of  the  spring  should  act  ex- 
actly along  the  axis  of  the  spring. 

This  is  of  especial  importance  when  the  spring  is  used  for 
making  accurate  measurements  of  forces,  as  in  the  steam-en- 
gine indicator,  in  spring  balances,  etc. 

Moreover,  it  is  generally  safer,  as  far  as  accuracy  is  con- 
cerned, to  use  a  helical  spring  in  tension  rather  than  in  com- 
pression, as  it  is  easier  to  make  sure  that  the  forces  act  along 


342  APPLIED   MECHANICS. 

the  axis  in  the  case  of  tension  than  in  the  case  of  compres- 
sion. 

Whichever  way  the  spring  is  used,  however,  provided  only 
the  two  opposing  forces  act  along  the  axis  of  the  spring,  the 
resistance  to  which  the  spring  is  subjected  is  mainly  torsion, 
inasmuch  as  the  amount  of  bending  is  very  slight. 

This  bending,  however,  we  will  neglect,  and  will  compute 
the  spring  as  a  case  of  pure  torsion,  the  same  notation  being 
used  as  before,  except  that  we  will  now  denote  by  R  the  radius 


of  the  spring,  and  we  shall  have 

M  =  LR,    M,  = 

and  now  formulae  (5),  (6),  (7),  and  (8)  become  applicable  to  a 
spring  made  of  round  wire,  and  formulae  (9)  and  (10),  (11)  and 
(12),  to  one  made\  of  rectangular  wire. 

We  must  bear  in  mind,  however,  that  x  denotes  the  length 
of  the  wire  composing  the  spring,  and  not  the  length  of  the 
spring,  d  and  <5\  now  denote  the  elongations  or  compressions 
of  the  spring. 

GENERAL   REMARKS. 

By  comparing  equations  (8)  and  (12),  it  will  be  seen 
that  if  a  spring  is  required  for  a  given  service,  its  volume 
and  hence  its  weight  must  be  50  per  cent  greater  if  made 
of  rectangular  than  if  made  of  round  wire.  Again,  it  is 
evident  that  when  the  kind  of  spring  required  is  given, 


SPRINGS.  343 


and  the  values  of  C  and  f  for  the  material  of  which  it  is  to 
be  made  are  known,  the  volume  and  hence  the  weight  01 
me  spring  depends  only  on  the  product  Pd^  and  that  as  soon 
as  P  and  rf,  are  given,  the  weight  of  the  spring  is  fixed  inde- 
pendently of  its  special  dimensions.  If,  however,  we  fix  any 
one  dimension  arbitrarily,  the  others  must  be  so  fixed  as  to 
satisfy  the  equations  already  given.  Next,  as  to  the  values  to 
be  used  for /and  C,  these  will  depend  upon  the  nature  of  the 
special  material  of  which  the  spring  is  made,  and  these  can 
only  be  determined  by  experiment.  Confining  ourselves  now 
to  the  case  of  steel  springs,  it  is  plain  that  /and  C  should  be 
values  corresponding  to  tempered  steel. 

As  an  example,  suppose  we  require  the  weight  of  a  helical 
spring,  which  is  to  bear  a  safe  load  of  10000  Ibs.  with  a  deflec- 
tion of  one  inch,  assuming  C  —  12600000  and/=  80000  Ibs. 
per  sq.  in.,  and  as  the  weight  of  the  steel  0.28  Ib.  per  cubic 
inch. 

From  formula  (8)  we  obtain 

2  X  12600000  X  10000  X  I 
V  —  —  39.4  cu.  in. 

80000  X  80000 

Hence  the  weight  of  the  spring  must  be  (39.4)  (0.28)  —  1 1  Ibs. 
We  may  use  either  a  single  spring  weighing  1 1  Ibs.,  or  else 
two  or  more  springs  either  side  by  side  or  in  a  nest,  whose  com- 
bined weight  is  u  Ibs.  Of  course  in  the  latter  case  they  must 
all  deflect  the  same  amount  under  the  portion  of  the  load 
which  each  one  is  expected  to  bear,  and  this  fact  must  be 
taken  into  account  in  proportioning  the  separate  springs  that 
compose  the  nest. 

FLAT   SPRINGS. 

Let  P,  L,  V,  tf,  and  tf,  have  the  same  meanings 'as  before, 
and  let 


344 


APPLIED  MECHANICS. 


f=  greatest  allowable  fibre  stress  for  tension  or  compres- 

sion : 

E  =  modulus  of  elasticity  for  tension  or  compression  ; 
/=  length  of  spring; 

Ml  =  maximum  bending-moment  under  load  P  ; 
M  =  maximum  bending-moment  under  load  L. 

Moreover,  the  sections  to  be  considered  are  all  rectangular, 
and  we  will  let  b  =  breadth  and  h  =  depth  at  the  section 
where  the  greatest  bending-moment  acts,  the  depth  being 
measured  parallel  to  the  load. 

Then  if  /  denote  the  moment  of  inertia  of  the  section  of 
greatest  bending-moment  about  its  neutral  axis,  we  shall  have 


12 

We  will  now  consider  six  cases  of  flat  springs,  and  vail  de- 
termine P,  #,  tfx  ,  and  V  for  each  case,  and  for  this  purpose  we 
only  need  to  apply  the  ordinary  rules  for  the  strength  and  de- 
flection of  beams. 

CASE  i.  Simple  rectangular  spring,  fixed  at  one  end  and 
loaded  at  the  other. 


Is  L 


(26) 


SPRINGS. 


345 


CASE  2.  Spring  of  uniform  depth  and  uniform  strength,  tri- 
angular in  plan,  fixed  at  one  end  and  loaded  at  the  other. 


•-**£ 

-ts  -f, 

A       f.1"  L 
S  =  (>W£' 


*,= 


_£!/. 


**' 


(^7) 
(28) 

(29) 


(30) 


CASE  3.  Spring  of  uniform  breadth  and  uniform  strength, 
parabolic  in  elevation,  fixed  at  one  end  and  loaded  at  the 
other. 


_ 

I ' 


- 


«-* 


(33) 


v, 


(34) 


CASE  4.  Compound  wagon  spring,  made  up  of  n  simple  rec- 
tangular springs  laid  one  above  the  other,  fixed  at  one  end  and 
loaded  at  the  other. 


346 


APPLIED   MECHANICS, 


Let  the  breadth  be  b,  and  the  depth  of  each  separate   layer 

be  h.     Then 

P  =  £/T>                      (35) 

'A 

^^^^ 

*  =  ~nw4'          (36) 

i 

-^ 

/?            2        •/                                           1       \ 

i. 

1      ^hE'                         ^' 

(38) 


CASE  5.  Compound  spring  composed  of  n  triangular  springs 
laid  one  above  the  other,  fixed  at  one  end  and  loaded  at  the 
other. 


r 


-_ 

~  nbtf  E' 


h  E' 


(39) 
(4o) 


.-.V=3i 


CASE  6.  This  case  differs  from  the  last  in  that  in  order  to 
economize  material  we  superpose  springs  of  different  lengths, 


347 


and  make  them  of  such  a  shape  that  by  the  action  of  a  single 
force  at  the  free  end  they  are  bent  in  arcs  of  circles  of  nearly 
or  exactly  the  same  radius. 
The  force  P  bends  the 
lowest  triangular  piece  AA 
in  the  arc  of  a  circle.     The 

length  of  this  piece  is  — . 


-  p 


-p 


n 

In  order  that  the  re- 
maining parallelopipedical 
portion  may  bend  into  an 
arc  of  the  same  circle  it  is 
necessary  that  it  should  have 

acting  on  it  a  uniform  bending-moment  throughout,  and  this 
is  attained  if  it  exerts  a  pressure  at  At  upon  the  succeeding 
spring  equal  to  the  force  P,  and  following  this  out  we  should 
find  that  the  entire  spring  would  bend  in  an  arc  of  a  circle. 

The  values  of  P,  d,  #x ,  and  Fare  correctly  expressed  for 
this  case  by  (39),  (40),  (41),  and  (42). 

For  any  flat  springs  which  are  supported  at  the  ends  and 
loaded  at  the  middle,  or  where  two  springs  are  fastened  to- 
gether, it  is  easy  to  compute,  by  means  of  the  formulae  already 
developed,  by  making  the  necessary  alterations,  the  quantities 
P,  <S  tfj ,  and  V,  and  this  will  be  left  to  the  student. 

COILED    SPRINGS    SUBJECTED    TO    TRANSVERSE    STRESS. 

Three  cases  of  coiled  springs  will  now  be  given  as  shown 
in  the  figures,  and  the  values  -of  P,  tf,  dlt  and  Fwill  be  deter- 
mined for  each. 

In  each  of  these  cases  let  R  be  the  leverage  of  the  load, 
and  let  GO  =  angle  turned  through  under  the  load.  Then  we 
may  observe  that  all  the  three  cases  are  cases  of  beams  sub- 
jected to  a  uniform  bending-moment  throughout  their  length, 
this  bending-moment  being  LR  for  load  L  and  PR  for  load  P. 


348 


APPLIED  MECHANICS. 


CASES  I  and  2.  Coiled  spring,  rectangular  in  section. 

*    "       ,  (41)  Illlllllllil 

Fig.  13. 

11?  L 


(43) 
(44) 


(45) 


E 
E 


CASE  3.  Coiled  spring,  cir- 
cular in  section. 


64 


(47) 
(48) 

(49) 


(46) 


TIME    OF    OSCILLATION    OF   A    SPRING. 

Since  in  any  spring  the  load  producing  any  displacement 
is  proportional  to  the  displacement,  it  follows  that  when  a 
spring  oscillates,  its  motion  is  harmonious. 


SPRINGS.  349 


Suppose  the  load  on  the  spring  to  be  Pt  and  hence  its  nor- 
mal displacement  to  be  d^.  Now  let  the  extreme  displacements 
on  the  two  sides  of  &l  be  &0  ,  and  the  force  producing  it  /,  so  that 
the  actual  displacement  varies  from  tfx  -f-  $0  to  tf:  —  $0  ,  and  the 
force  acting  varies  from  P  -\-  p  to  P  —  /. 

Now.  from  the  properties  of  the  spring  we  must  have 


Moreover,  in  the  case  of  harmonic   motion  the  maximum 

Wofr 
value  of  the  force  acting  is  -  (see  p.    104).     But  the  load 

o 

oscillating  is  P  instead  of  W,  and  the  extreme  displacement  is 
#0  instead  of  r. 
Hence  we  have 


(52) 


(53) 
Hence  the  time  of  a  double  oscillation 

(54) 
* 


35O  APPLIED   MECHANICS. 


CHAPTER    VII. 

STRENGTH    OF  MATERIALS   AS  DETERMINED    BY 
EXPERIMENT. 

§  216.  General  Remarks.  —  Whatever  computations  are 
made  to  determine  the  form  and  dimensions  of  pieces  that 
are  to  resist  stress  and  strain,  must,  if  they  are  to  have  any 
practical  value,  be  based  upon  experiments  made  upon  the  ma- 
terials themselves. 

The  most  valuable  experiments  in  any  given  case,  whenever 
the  results  of  such  experiments  are  available,  are  those  made 
upon  pieces  of  the  same  quality,  size,  and  form  as  those  to 
which  the  results  are  to  be  applied,  and  under  conditions  en- 
tirely similar  to  those  to  which  the  pieces  are  subjected  in 
actual  practice. 

It  is  not  very  often  that  the  results  of  such  experiments  are 
available  ;  and  hence  we  must,  in  general,  make  use  of  such 
tests  as  have  been  or  can  be  made,  and  from  them  determine 
the  strength  of  the  pieces  in  actual  use  by  computation,  making 
good  use  of  our  judgment. 

As  time  goes  on,  and  experimental  science  advances,  a 
greater  number  of  the  conditions  that  exist  in  actual  practice 
are  introduced  into\he  experiments  ;  and  hence  the  reliability 
of  the  experimental  results,  and  their  applicability  to  practical 
cases,  are  increased.  Nevertheless,  it  is  necessary  to  use  the 
utmost  caution  when  applying  the  results  of  the  experiments  to 
cases  where  the  conditions  are  different  from  those  under  which 
the  experiments  were  made.  An  attempt  will  be  made  in  this 


GENERAL   REMARKS.  351 

chapter  to  give  an  account  of  the  most  important  results  of 
experiment  on  the  strength  of  materials,  and  to  explain  the 
modes  of  using  the  results  that  are  now  employed.  As  to 
the  way  in  which  the  experiments  have  generally  been  carried 
on,  we  may  observe  :  — 

i°.  In  by  far  the  greater  number  of  cases,  the  test  pieces  have 
been  very  much  smaller  than  the  pieces  to  be  used.  Indeed, 
it  is  only  of  late  years  that  the  importance  of  testing  full-size 
pieces  has  been  recognized ;  and  hence  most  of  the  experiments 
upon  such  pieces  are  of  very  recent  date. 

2°.  In  the  greater  part  of  the  experiments  that  have  been 
made,  the  phenomena  observed  have  been  those  that  occur 
during  the  application  of  the  load  for  a  short  time  only ;  very 
little  having  been  done  by  way  of  determining  the  behavior  of 
the  pieces  under  a  long-continued  action  of  the  load,  or  under 
repeated  applications  of  the  load,  such  as  occur  in  practice. 

3°.  Very  few  experiments  have  been  made  on  the  effect  of 
applying  two  kinds  of  stress  simultaneously,  as  tension  and 
bending,  or  twisting  and  bending,  or  on  applying  stresses  of 
opposite  kinds,  as  tension  and  compression,  successively. 

4°.  The  tests  thus  far  made  have  had  for  their  object  more 

frequently  to  determine  the    breaking-strength  of    the  piece. 

-Next  to  this,  the  subject  most  frequently  experimented  upon 

has  been  the  limit  of  elasticity ;  and  less  has  been  done  by  way 

of  determining  the  modulus  of  elasticity,  and  other  matters. 

5°.  The  fact  that  the  breaking-strength  alone  is  not  a  suffi- 
cient criterion  by  which  to  determine  the  suitability  of  a  mate- 
rial for  use  in  construction,  has  been  recognized  only  by  the 
later  experimenters. 

6°.  In  order  to  understand  what  is  meant  by  "  the  limit  of 
elasticity,"  we  must  observe,  that,  if  a  small  load  be  applied  to 
the  piece  under  test,  and  then  removed,  the  deformation  or  dis- 
tortion caused  by  the  application  of  the  load  apparently  van- 
ishes, and  the  piece  resumes  its  original  form  and  dimensions 


352  APPLIED   MECHANICS. 

on  the  removal  of  the  load  ;  in  other  words,  no  permanent  set 
takes  place.  When  the  load,  however,  is  increased  beyond  a 
certain  point,  the  piece  under  test  does  not  return  entirely  to 
its  original  dimensions  on  the  removal  of  the  load,  but  retains  a 
certain  permanent  set. 

The  load  upon  the  application  of  which  permanent  set 
apparently  begins,  is  called  the  limit  of  elasticity,  and  is  found 
by  experiment  to  be  at  about  one-third  the  breaking-weight  in 
iron,  and  from  one-third  upwards  in  steel,  sometimes  reaching 
nearly  three-fourths. 

Experiments  show,  however,  that  even  a  very  small  load  will 
produce  a  permanent  set,  and  that  the  apparent  return  of  the 
piece  to  its  original  dimensions  upon  the  removal  of  the  load  is 
only  due  to  the  want  of  delicacy  in  the  measuring-instruments 
that  have  been  used  in  the  tests.  A  better  definition  of  the 
limit  of  elasticity  would  therefore  be,  that  load  upon  the  appli- 
cation of  which  the  permanent  set  begins  to  be  noticeable,  with 
such  rough  means  of  measuring  as  a  pair  of  dividers. 

7°.  It  has  often  been  assumed,  that,  if  the  load  applied  to 
the  piece  in  practice  exceeded  the  elastic  limit,  the  piece  would 
be  permanently  injured  in  its  properties  for  resisting  stress,  and 
that  the  deformation  and  injury  would  continue  increasing,  until 
eventually  fracture  would  occur.  It  has  been  proved  experimen- 
tally, however,  that  it  is  sometimes  advantageous  to  apply  once 
a  load  to  a  piece  somewhat  greater  than  the  elastic  limit,  and 
that  by  this  means  the  elastic  limit  is  increased.  This  process 
of  using  up  a  part  of  the  elasticity  of  the  piece  cannot  continue 
indefinitely,  and  the  data  to  show  how  far  it  can  be  advanta- 
geously carried  are  but  few. 

8°.  The  determination  of  the  modulus  of  elasticity,  which 
has  been  defined  (§  167)  as  the  ratio  of  the  stress  to  the  strain, 
is  a  very  important  matter ;  as  it  gives  us  the  means  of  com- 
puting the  deformation  under  any  given  load,  and  thus  deter- 
mining the  safe  load  by  prescribing  the  greatest  deformation  to 


GENERAL   REMARKS.  353 


be  allowed,  rather  than  by  prescribing  that  the  safe  load  shall 
be  a  certain  fraction  of  the  breaking-load. 

9°.  Other  important  matters  which  guide  us  in  judging  of 
the  suitability  of  a  piece  for  the  purpose  to  which  it  is  to  be 
applied,  are,  the  appearance  of  its  fracture,  its  density,  its  homo- 
geneity, its  composition,  and  the  care  taken  in  its  manufacture, 
or  the  circumstances  of  its  growth  and  seasoning  if  it  is  wood, 
also  its  brittleness,  hardness,  malleability,  ductility,  the  amount 
of  warning  it  gives  before  giving  way,  etc. 

10°.  From  such  data  as  we  have  furnished  to  us  by  experi- 
ment, we  decide,  to  the  best  of  our  ability,  as  to  the  suitability 
of  the  piece  for  the  use  for  which  it  is  intended,  and  also  as  to 
the  amount  it  will  safely  bear,  when  we  know  its  dimensions, 
or  the  proper  dimensions  to  bear  safely  the  required  stress. 

11°.  Tests  have  been  made  on  tension,  compression,  shear- 
ing, transverse  and  torsional  strength,  of  the  different  materials 
used  in  construction,  especially  cast-iron,  wrought-iron,  steel, 
and  wood,  also  copper  and  other  metals ;  but  the  tests  on  ten- 
sile strength  are  by  far  the  most  numerous  in  the  case  of  iron. 

§  217.  Cast-iron.  —  Cast-iron  is  a  combination  of  iron  with 
2  per  cent  to  6  per  cent  of  carbon.  The  large  amount  of  carbon 
which  it  contains  is  its  distinguishing  feature,  and  determines 
its  behavior  in  most  respects. 

Pig-iron  is  the  result  of  the  first  smelting,  being  obtained 
directly  from  the  smelting-furnace.  The  ore  and  fuel  are  put 
into  the  furnace,  together  with  a  flux,  which  is  of  a  calcareous 
nature  when  the  ore  is  argillaceous,  or  which  contains  clay 
when  the  ore  is  cal<^reous.  The  mass  is  brought  to  a  high 
heat,  a  strong  blast  of  air  being  introduced.  The  mass  is  thus 
melted ;  the  fluid  iron  settling  to  the  bottom,  while  slag,  which 
is  the  result  of  the  combination  of  the  flux  with  the  impurities 
of  the  ore,  rises  to  the  top.  The  iron  is  drawn  off  in  the  liquid 
state,  and  run  into  moulds,  the  result  being  pig-iron. 

The  result  of  this  first  melting  is  very  rarely  used  for  any 


354  APPLIED   MECHANICS. 

casting ;  but  the  pig-iron  is  usually  re-melted  in  a  cupola  furnace 
before  being  used,  the  result  of  this  re-melting  being  the  ordi- 
nary cast-iron  of  commerce. 

The  pig-iron  is  divided  into  classes,  according  to  the  purpose 
for  which  it  is  intended,  and  the  amount  of  carbon  it  contains. 

Those  pigs  that  have  a  considerable  amount  of  carbon  in  me- 
chanical mixture,  and  show  a  gray  color  on  being  fractured,  are 
used  by  the  founder  to  melt  over,  and  make  cast-iron.  This  is 
called  "foundry  iron,"  and  is  divided  into  foundry  iron  Nos.  I, 
2,  and  3,  from  which  are  subsequently  made  gray  cast-iron 
Nos.  i,  2,  and  3.  When  there  is  less  carbon,  it  is  sometimes 
called  "foundry  No.  4,"  etc.  ;  but  it  is  only  used  to  make 
wrought-iron  of  an  inferior  quality. 

Those  pigs  which  are  to  be  used  in  making  wrought-iron 
and  steel,  and  which  have  been  fused  at  a  low  heat  and  with 
little  fuel,  are  called  "forge-iron." 

Cast-iron  is  of  two  kinds,  white  cast-iron  and  gray  cast-iron. 
The  first  is  a  chemical  compound  of  iron  with  2  per  cent  to  6  per 
cent  of  carbon,  almost  all  of  the  carbon  being  chemically  com- 
bined with  the  iron.  The  second,  or  gray  cast-iron,  contains 
part  of  the  carbon  in  chemical  combination,  and  the  remainder 
in  the  state  of  graphite  mechanically  mixed  with  the  iron. 

Gray  Cast-Iron  is  divided  into  three  classes,  known  respec- 
tively as  Nos.  i,  2,  and  3. 

No.  I  contains  the  largest  amount  of  carbon  in  mechanical 
mixture,  the  effect  of  which  is  to  render  it  soft  and  fusible, 
though  not  as  strong  as  Nos.  2  and  3.  It  is,  therefore,  very 
suitable  for  making  castings  where  precision  in  form  is  a  desid- 
eratum, as  its  fusibility  causes  it  to  fill  the  mould  well.  It  is 
not  as  suitable,  however,  where  strength  is  required. 

No.  2  is  that  which  is  most  suitable  for  use  in  construction, 
as  it  is  stronger  than  No.  i,  and  not  so  soft. 

No.  J,  on  the  other  hand,  contains  the  smallest  amount  of 
carbon  in  the  graphitic  form,  and  is,  hence,  harder  and  more 


CAST-IRON.  355 


brittle.  It  is  suited,  therefore,  only  for  the  massive  and  heavy 
parts  of  machinery. 

White  Cast-Iron  contains  hardly  any  free  carbon.  It  is  of  two 
kinds,  granular  and  crystalline.  The  crystalline  variety  is  of 
no  use  in  construction  :  it  is  hard  and  very  brittle.  The  granu- 
lar variety  is  also  unsuitable  for  use  in  construction,  but  forms 
the  hard  skin  on  the  surface  of  a  piece  that  has  been  chilled. 

As  to  the  adaptability  of  cast-iron  to  construction,  it  pre- 
sents certain  advantages  and  certain  disadvantages.  It  is  the 
cheapest  form  of  iron.  It  is  easy  to  give  it  any  desired  form. 
It  resists  oxidation  better  than  either  wrought-iron  or  steel.  It 
has  a  very  high  compressive  strength.  On  the  other  hand,  its 
tensile  strength  is  comparatively  small,  averaging,  in  common 
varieties,  15000  pounds  per  square  inch,  or  thereabouts.  It 
cannot  be  riveted;  or  welded  by  forging.  It  is  brittle,  break- 
ing off  without  giving  much  warning,  and  stretching  but  little 
before  giving  way.  It  is  liable  to  hidden  and  small  surface 
defects  and  air-bubbles,  which  render  its  strength  somewhat 
doubtful.  It  is  also  liable  to  absorb  impurities  from  the  fuel  or 
flux  in  the  furnace,  the  most  injurious  being  sulphur  and  phos- 
phorus ;  the  effect  of  the  former  being  to  produce  red  short- 
ness, or  brittleness  when  hot,  and  that  of  the  latter  to  produce 
cold  shortness,  or  brittleness  when  cold. 

Another  very  serious  drawback  in  the  use  of  cast-iron  in  con- 
struction is  its  liability  to  initial  strains  from  the  inequality 
in  cooling.  Thus,  if  one  part  of  the  casting  is  very  thin  and 
another  very  thick,  the  thin  part  cools  first,  and,  in  cooling,  con- 
tracts ;  and  the  thick  part,  cooling  afterwards,  causes  stresses 
in  the  thin  part,  which  may  be  sufficient  to  break  it,  or,  if  not, 
there  may  be  so  much  stress  established,  that  but  little  more 
will  break  it.  Thus,  the  change  of  temperature  from  summer 
to  winter  is  sometimes  sufficient  to  break  the  arms  of  a  pulley 
from  off  the  rim.  Its  quality  depends  largely  upon  its  composi- 
tion and  its  density. 


356  APPLIED  MECHANICS. 

The  fracture  should  be  of  a  bluish-gray  color,  and  close- 
grained  texture,  with  considerable  metallic  lustre  if  the  iron  is 
of  good  quality.  If  the  fracture  is  mottled,  with  patches  of 
darker  or  lighter  iron,  or  crystalline  spots,  it  is  an  indication 
of  unsoundness,  especially  so  if  there  are  air-bubbles. 

It  is  not  well  adapted  to  bear  tension,  on  account  of  its  low 
tensile  strength,  and  also  on  account  of  its  brittleness  and 
treacherousness. 

In  former  times  it  was  extensively  used  for  iron  beams  to 
bear  a  transverse  load,  but  has  now  been  almost  entirely  super- 
seded by  wrought-iron  in  this  regard. 

It  is  still  used  for  columns  and  posts  in  buildings,  on  ac- 
count of  its  high  compressive  strength  ;  but,  in  cases  where  the 
length  of  the  column  is  so  great  as  to  cause  it  to  give  way  by 
bending,  as  in  bridge  columns,  its  use  has  been  almost  wholly 
abandoned,  and  wrought-iron  and  steel  are  taking  its  place.  In 
the  case  of  bridge  columns,  it  is  also  necessary  to  use  a  metal 
which  can  easily  be  riveted,  and  to  which  the  other  members 
can  be  readily  attached;  and  wrought  iron  is  more  suitable  than 
cast  for  this  purpose  :  also  another  reason  is,  that  wrought-iron 
and  steel  are  much  better  suited  to  resist  shocks  than  cast-iron. 

In  machinery,  it  is  used  in  all  those  parts  where  weight, 
mass,  or  form  is  of  more  importance  than  strength,  as  in  the 
frames  and  bed-plates  of  machines,  also  for  hangers,  pulleys, 
and  gear-wheels. 

Cast-iron  is  also  used  for  water-mains  where  great  pressure 
is  to  be  resisted,  also  in  hydraulic  presses,  and  in  heavy  ord- 
nance. For  shafting,  wrought-iron  has  taken  its  place,  and  so 
also  for  the  shells  of  steam-boilers,  with  the  exception  of  some 
sectional  boilers,  partly  on  account  of  its  low  tensile  strength 
and  general  untrustworthiness,  but  especially  because  of  its 
liability  to  give  way  without  warning  when  subjected  to  the 
sudden  expansions  and  contractions  which  it  would  have  to 
undergo  if  used  in  a  steam-boiler. 


CAST-IRON.  357 


Malleable  Cast-Iron.  —  When  a  casting  is  to  be  made  in.  a 
rather  intricate  form,  it  is  frequently  the  custom  to  malleableize 
the  cast-iron.  This  is  done  by  heating  it  to  a  bright-red  heat  in 
an  annealing  oven,  in  powdered  hematite  ore,  with  a  suitable 
flux.  By  this  process  a  part  of  the  carbon  is  removed,  and  the 
result  is  —  provided  the  casting  is  not  large  —  a  product  that 
can  be  hammered  into  any  desired  shape  when  cold,  but  is  very 
brittle  when  hot.  It  is  used  in  cases  where  toughness  is  re- 
quired, together  with  the  possession  of  an  intricate  form. 

§  218.  Tensile  and  Compressive  Strength  of  Cast-Iron. 
— A  list  of  references  to  some  of  the  principal  experimental 
works  on  the  strength  and  elasticity  of  cast-iron  will  be  given. 

i°.  Eaton  Hodgkinson  :  (a)  Report  of  the  Commissioners  on  the 
Application  of  Iron  to  Railway  Structures. 

(b)  London  Philosophical  Transactions.      1840. 

(c)  Experimental  Researches  on  the  Strength  and  other  Prop- 
erties of  Cast-Iron.      1846. 

2°.  W.  H.  Barlow  :  Barlow's  Strength  of  Materials. 

3°.  Sir  William  Fairbairn  :  On  the  Application  of  Cast  and  Wrought 
Iron  to  Building  Purposes. 

4°.  Major  Wade  (U.SA.)  :  Report  of  the  Ordnance  Department 
on  the  Experiments  on  Metals  for  Cannon.  1856. 

5°.  Capt.  T.  J.  Rodman  :   Experiments  on  Metals  for  Cannon. 

6°.  Col.  Rosset:  Resistenza  dei  Principali  Metalli  daBocchidi  Fuoco. 

7°.  Report  for  1887,  of  the  Tests  of  Metals,  made  on  the  Govern- 
ment Testing  Machine  at  Watertown  Arsenal. 

8°.  Proceedings  of  the  Am.  Soc.  of  Mech..  Engineers  for  1889, 
page  187  et  seq. 

9°.  Bauschinger  :  Mittheilungen  aus  dem  Mech.  Tech.  Lab.  Miin- 
chen.  Heft  12,  1885,  ar>d  Heft  15,  1887. 

10°.  Technology  Quarterly,  October,  1888,  page  12  et  seq. 

Hodgkinson  and  the  other  members  of  the  commission 
made  tests  to  compare  the  strength  of  iron  from  different  parts 
of  the  British  Isles. 


353 


APPLIED   MECHANICS. 


The  following  table  gives  the  tensile  and  also  the  crushing 
strength  per  square  inch  of  the  different  kinds  tested  :  — 


Description  of  the  Iron. 


Tensile  Strength 

per 

Square  Inch, 
in  Ibs. 


Height  of 
Specimen   for 
Compression, 

in  inches. 


Crushing-Strength 

per  Square  Inch, 

in  Ibs. 


Monies  Stirling's  (second  quality), 
Morries  Stirling's  (third  quality)   . 


Lowmoor,  No.  i 12694 

Lowmoor,  No.  2 *5458 

Clyde,  No.  i 16125 

Clyde,  No.  2 17807 

Clyde,  No.  3 23468 

Blaenavon,  No.   i 

Blaenavon,  No.  2  (first  sample)      .  16724 

Blaenavon,  No.  2  (second  sample),  14291 

Calder,  No.  i  .........  T3735 

Coltness,  No.  3 15278 

Brymbo,  No.  i 14426 

Brymbo,  No.  3 I55°8 

Bowling,  No.  2 

Ystalyfera,  No.  2  (anthracite)  .  . 

Yniscedwyn,  No.  i  (anthracite)   .  I3952 

Yniscedwyn,  No.  2  (anthracite)   .  13348 


25764 
23461 


I.! 
14 
ft 


1 4 

u 

4 


64534 
56445 
99525 


88741 

109992 

102030 

107197 

104881 

90860 

80561 

117605 

102408 

68559 

68532 

72193 

75983 
100180 
101831 

74815 
75678 

76133 

76958 
76132 
73984 
99926 
95559 

83509 
78659 
77124 
75369 

I25333 
"9457 


1 553 
129876 


TENSILE,  ETC.,  STRENGTH  OF  CAST-IRON.  359 

The  specimens  used  for  determining  the  tensile  strength 
were  all  cruciform  in  section ;  those  for  determining  the  crush- 
ing-strength were  cylinders,  with  a  diameter  of  |  inch.  The 
machine  used  was  simply  a  single  lever,  with  a  scale-pan  at  the 
extreme  end,  in  which  weights  were  placed.  There  was  no 
provision  for  taking  up  the  stretch  of  the  specimen,  and  there- 
fore the  specimens  were  all  short. 

The  average  tensile  strength  of  these  specimens,  omitting 
the  Stirling  metal,  was  15298  Ibs.  per  square  inch;  and  the 
average  compressive  strength  was  82296  Ibs.  per  square  inch. 

It  will  be  observed,  that,  omitting  the  Stirling  iron,  which 
is  a  mixture  of  cast  and  wrought  iron,  the  tensile  strength 
ranged  from  12694  to  23468  Ibs.  per  square  inch,  and  the  com- 
pressive strength,  from  56445  to  117605  Ibs.  per  square  inch. 
The  reader  will  doubtless  observe,  that,  as  a  rule,  the  crushing- 
strengths  obtained  from  the  longer  specimens  were  less  than 
those  obtained  with  the  shorter  ones ;  but  this,  it  seems  to  the 
writer,  is  probably  due  to  the  nature  of  the  testing-machine, 
and  not  to  any  bending  in  the  specimen  due  to  its  length. 

Mr.  Hodgkinson  next  proceeded  to  compare  the  tensile 
strength  of  specimens  cruciform  in  section,  with  that  of  speci- 
mens circular  in  section  and  of  about  the  same  area.  He  found 
but  little  difference ;  and  this  could  readily  be  accounted  for  by 
the  fact,  that,  the  perimeter  of  the  cruciform  one  being  greater, 
the  proportion  of  hard  skin  would  be  greater  in  the  cruciform 
than  in  the  circular,  the  effect  of  this  being,  perhaps,  partially 
counteracted  by  some  little  initial  stress,  on  account  of  unequal 
cooling  of  the  different  parts. 

Hodgkinson  also  made  a  few  experiments  to  determine  the 
laws  of  extension  of  cast-iron,  and  for  this  purpose  used  rods 
10  feet  long  and  I  square  inch  in  section.  The  table  of  average 
results  is  the  following  :  — 


360 


APPLIED   MECHANICS. 


RESULTS    OF    NINE    TENSILE    TESTS. 


Weights  laid 
on, 
in  Ibs. 

Extensions, 
in 
inches. 

Sets, 
in  inches. 

Strains,  in 
fractions  of 
the  length. 

Sets,  in 
fractions  of 
the  length. 

Ratio  of 
Weight  to 
Extensions 

Modulus 
of 
Elasticity. 

105377 

0.0090 

_ 

O.OOOO7 

_ 

II7086 

14050320 

1580.65 

0.0137 

O.OOO22 

O.OOOI  I 

O.OOOOOlS 

H5I3I 

13815720 

2  [07.54 

0.0186 

O.OOO55 

0.00016 

0.0000046 

113309 

13597080 

3I6I.3I 

0.0287 

O.OOIO7 

0.00024 

0.0000089 

110150 

13218000 

4215.08 

0.0391 

0.00175 

0.00033 

0.0000146 

107803 

I  2936360 

5268.85 

0.0500 

0.00265 

O.OOO42 

O.OOOO22I 

105377 

12645240 

6322.62 

0.0613 

0.00372 

0.00051 

O.OOOO3IO 

103142 

12377040 

7376.39 

0.0734 

0.00517 

O.OOO6I 

0.0000431 

100496 

12059520 

8430.16 

0.0859 

0.00664 

O.OOO72 

0.0000553 

98139 

11776680 

9483.94 

0.0995 

0.00844 

0.00083 

O.OOOO7O3 

95316 

11437920 

10537.71 

0.1136 

O.OIO62 

O.OOO95 

O.OOOO885 

92762 

11131440 

11591.48 

0.1283 

0.01306 

0.00107 

0.0001088 

90347 

10841640 

12645.25 

0.1448 

0.01609 

O.OOI2I 

O.OOOI34I 

87329 

10479480 

1  3699-83 

0.1668 

0.02097 

O.OOI39 

0.0001748 

82133 

9855960 

14793.10 

0.1859 

0.02410 

O.OOI55 

0.0002008 

79576 

9549120 

RESULTS    OF    EIGHT    COMPRESSIVE    TESTS. 


Weights  laid 
on, 
in  Ibs. 

Compres- 
sions, 
in  inches. 

Sets, 
in  inches. 

Strains,  in 
fractions  of 
the  length. 

Sets,  in 
fractions  of 
the  length. 

Ratio  of 
Weight  to 
Compres- 
sions. 

Modulus 
of 
Elasticity. 

2064.75 

0.01875 

0.00047 

O.OOOI6 

O.OOOOO39 

IIOI20 

13214400 

4129.49 

0.03878 

0.00226 

O.OOO32 

O.OOOOlSS 

106485 

12778200 

6194.24 

0.05978 

0.00400 

O.OOO5O 

0.0000333 

103617 

12434040 

8258.98 

0.07879 

0.00645 

0.00066 

0.0000538 

104823 

12578760 

1032373 

0.09944 

0.00847 

0.00083 

0.0000706 

103819 

12458280 

12388.48 

0.12030 

0.01088 

O.OOIOO 

0.0000907 

102980 

12357600 

14453-22 

0.14163 

0.01405 

0.00118 

O.OOOII7I 

IO2O49 

12245880 

16517.97 

0.16338 

O.OI7I2 

0.00136 

0.0001427 

IOIIO2 

12132240 

18582.71 

0.18505 

0.02051 

0.00154 

0.0001709 

100420 

12050400 

20647.46 

0.20624 

0.02484 

0.00172 

O.OOO2O7O 

I001I4 

12013680 

24776.95 

0.24961 

0.03220 

0.00208 

0.0002683 

99263 

11911560 

28906.45 

0.29699 

0.04300 

0.00247 

0.0003583 

9733  l 

116/9720 

33030.80 

0-35341 

0.06096 

0.00295 

0.0005080 

93463 

11215560 

RESULTS   OF   TESTS. 


36i 


These  tables  show  that  the  modulus  of  elasticity  of  cast- 
iron  varies  with  the  load,  growing  gradually  smaller  as  the  load 
increases. 

The  following  table  enables  us  to  compare  the  modulus  of 
elasticity  for  tension  with  that  for  compression  under  nearly  the 
same  load. 


Tension. 

Compression. 

Load 

Modulus  of 

Load 

Modulus  of 

per  Square 
Inch. 

Elasticity. 

per  Square 
Inch. 

Elasticity. 

2107 

13597080 

2064 

13214400 

4215 

12936360 

4129 

12778200 

6322 

12377040 

6194 

12434000 

8430 

11776680 

8258 

12578760 

10537 

11131440 

10323 

12458280 

12645 

10479480 

12388 

12357600 

H793 

9549120 

J4453 

12245880 

This  table  shows,  that,  with  moderate  loads,  the  modulus  of 
elasticity  for  tension  of  cast-iron  does  not  differ  materially  from 
that  for  compression,  and  that  the  difference  increases  as  the 
load  becomes  greater. 

This  fact  is  one  of  very  considerable  importance,  inasmuch 
as  it  is  one  of  the  fundamental  assumptions  in  the  common 
theory  of  beams  ;  and  we  thus  see  that  experiment  justifies  our 
making  this  assumption  for  cast-iron  whenever  the  load  is  nof 
excessive.  The  justification  is  even  greater  with  wrought-iron 
and  steel. 

The  gradual  decrease  of  the  modulus  of  elasticity  with  the 
increase  of  load  shows  that  Hooke's  law,  "  Ut  tensio  sic  vis" 
(the  stress  is  proportional  to  the  strain),  does  not  hold  true  in 


APPLIED   MECHANICS. 


cast-iron  ;  nevertheless,  it  is  more  nearly  true  for  moderate  loads 
than  for  larger  ones. 

The  experiments  of  Major  Wade  of  the  United-States  army, 
most  of  which  were  made  at  the  South-Boston  Iron  Foundry, 
are  recorded  in  the  Report  of  the  Ordnance  Department  on 
metals  for  cannon,  printed  in  1856.  The  Report  contains 
experiments  in  regard  to  the  effect  of  keeping  the  iron  in  a 
state  of  fusion  for  a  long  time,  and  also  on  the  effect  of  suc- 
cessive re-meltings  upon  the  quality  of  the  iron  ;  and  he  says,  — 

"  It  is  found  that  the  same  iron  is  greatly  improved  in  qual- 
ity by  retaining  it  in  the  furnace  after  it  is  melted  for  consider- 
able periods  of  time,  and  that  it  is  further  improved  by  casting 
it  into  pigs,  and  again  re-melting  it.  But  it  is  known  also  that  a 
continuation  of  this  process  will  ultimately  impair  the  tenacity 
of  the  iron,  and  render  it  wholly  unfit  for  use.  It  is  found  also 
that  the  different  kinds  of  iron  require  a  different  kind  of  treat- 
ment to  produce  the  best  effect.  The  breaking-instrument  en- 
ables one  to  ascertain  the  effect  produced  by  these  processes  in 
all  their  several  stages  of  progress,  and  to  decide  on  that  which 
is  found  most  suitable  for  making  guns  of  the  best  quality." 

On  p.  279  of  the  Report  is  given  the  following  experiment 
in  this  regard  :  — 

"The  first  sample  was  a  rough, 
crude  pig  of  grade  No.  i,  Green- 
wood iron. 

"  The  second,  third,  and  fourth 
are  the  same  iron,  cast  and  cooled 
in  like  manner,  and  differ  only  in 
t^ie  number  of  times  melted  ;  and 
they  exhibit  the  changes  effected 
in  the  strength  by  the  repeated 
meltings  only.  The  fifth  sample 
is  from  the  same  melting  as  the  fourth,  from  which  it  differs 
only  in  cooling,  being  cast  in  a  large  mass,  and  cooled  slowly." 


Fusion. 

Tensile 
Strength. 

Pig    - 

ist 

14000 

(2d 

20900 

Bars  . 

3d 

30229 

(.4th 

357S6 

Head, 

4th 

33724 

TENSILE,   ETC.,   STRENGTH  OF  CAST-IRON. 


363 


The  compressive  strength  of  cast-iron  tested  by  Major 
Wade  varied  from  84500  Ibs.  per  square  inch  to  175000  Ibs.  per 
square  inch.  The  following  table  of  results  of  a  number  of 
tests  of  metal  taken  from  different  cannon,  gives  the  average 
tensile  strength,  specific  gravity,  and  proportion  of  carbon  in 
the  lot  of  specimens  examined  :  — 


Tensile 

Specific 

Strength, 
1V>« 

Total 

Combined 

Allotropic 

Gravity. 

per  Square 

Carbon. 

Carbon. 

Carbon. 

Inch. 

i  st-  class  guns, 

7.204 

28805 

0.0384 

0.0178 

0.0206 

2d-class  guns, 

7-154 

24767 

0.0376 

0.0146 

0.0230 

3d-class  guns, 

7.087 

20148 

0.0365 

0.0082 

0.0283 

It  will  be  noticed  that  an  increase  in  tensile  strength  is 
generally  accompanied  by  an  increase  in  specific  gravity. 
Thus  another  lot  of  iron  gave  the  following  results  :  — 


Tensile 

Strength, 

Specific 

Ibs.,  per 

Gravity. 

Square  Inch. 

Mean 

27232 

7.302 

Least 

22402 

7.163 

Greatest, 

31027 

7.402 

He  also   experimented  on  the  difference  between  hot  and 
cold  blast  iron,  and  recommends  decidedly  the  cold  blast. 


APPLIED   MECHANICS. 


With  the  hot  blast,  unless  the  materials  —  i.e.,  the  ore,  fuel, 
and  flux  —  are  very  pure,  more  of  the  impurities  are  fused,  and 
combine  with  the  iron  ;  whereas  the  iron  can  be  made  much 
more  rapidly  by  its  use.  Indeed,  very  little  cold-blast  charcoal 
iron  is  made  at  the  present  time ;  i-e.,  iron  where  the  blast  is 
cold,  and  where  charcoal  is  the  only  fuel  used. 

The  rules  of  the  Ordnance  Department  of  the  United 
States  require  that  all  cast-iron  which  is  used  for  cannon  shall 
have  at  least  a  tensile  strength  of  30000  Ibs.  per  square  in.ch. 

The  specimens  used  for  testing  the  tensile  strength  of  cast- 
iron  have  generally  been  made  with  shoulders  :  and  the  smallest 
part  has  had,  in  most  cases,  no  length ;  and  the  specimen  has 
had,  therefore,  ve.ry  little  opportunity  to  stretch. 

Colonel  Rosset,  of  the  Arsenal  at  Turin,  made  a  series  of 
experiments  upon  the  influence  of  the  shape  of  the  specimen 
upon  the  tensile  strength.  For  this  purpose  he  used  specimens 
with  shoulders ;  and,  among  other  tests,  he  compared  the 
strength  of  the  same  iron  by  using  specimens  the  lengths  of 
whose  smallest  parts  were  respectively  I  metre,  30  millimetres, 
and  o  millimetres,  with  the  following  results  :  — 


Length  of  Specimen. 

Tensile  Strength,  in  Ibs.,  per  Square  Inch. 

ist  Cannon. 

2d  Cannon. 

3d  Cannon. 

i  metre      .     . 
30  millimetres  . 
o  millimetres  . 

31291 
32571 
33993 

25601 

34562 
36411 

28019 
30011 
30011 

It  will  thus  be  seen,  that,  before  we  can  decide  upon  the 
quality  of  cast-iron  as  affected  by  the  tensile  strength,  it  is 
necessary  to  know  the  length  of  that  part  of  the  specimen 


TENSILE,   ETC.,  STRENGTH  OF  CAST-IRON. 


565 


which  has  the  smallest  area.  Colonel  Rosset's  tests  of  cast> 
iron  were  almost  entirely  confined  to  high-grade  irons,  suitable 
to  use  in  cannons. 

He  deduced,  for  mean  value  of  the  modulus  of  elasticity  of 
the  specimens  I  metre  in  length,  20419658  Ibs.  per  square  inch  : 
this,  of  course,  is  a  modulus  only  adapted  to  these  high  grades, 
and  is  not  applicable  to  common  cast-iron. 

Before  proceeding  to  iron  columns,  the  values  given  for  the 
tensile  and  compressive  strength,  and  modulus  of  elasticity, 
of  cast-iron,  in  Ibs.,  per  square  inch,  given  by  Rankine  and 
Weisbach,  will  be  stated;  which  values  are  copied  by  many 
handbooks : — 


Tensile 
Strength. 

Compressive 
Strength. 

Modulus  of 
Elasticity. 

Professor  Rankine    . 

Various  qualities, 

(  13400 

to 

/     82OOO 
to 

f  I4OOOOOO 
to 

Professor  Weisbach, 

Average  .... 
Average  .... 

(  29OOO 
16500 
18500 

C  145000 

1  1  2OOO 

(  22900000 
17000000 
14220000 

§219.  Cast-Iron  Columns.  Hodgkinsoris  Experiments. — 
The  high  compressive  strength  of  cast-iron  would  seem  to  ren- 
der it  a  very  suitable  material  for  all  cases  of  columns  or  struts. 

Nevertheless,  its  use  for  the  compression  members  of  bridge 
and  roof  trusses  has  now  been  almost  entirely  abandoned;  and 
wrought-iron  has  taken  its  place  for  these  purposes,  —  partly 
because  these  struts  are  very  long,  and,  if  of  cast-iron,  might, 
on  account  of  the  bending,  bring  into  play  its  tensile  strength ; 
and  partly  because  wrought-iron  and  steel  can  be  easily  riveted, 
joined  to  other  pieces,  and  repaired,  whereas  cast-iron  cannot 
so  easily. 

Cast-iron  is  still  in  use  very  extensively  for  the  columns  of 


366  APPLIED  MECHANICS. 

buildings,  as  these  are  generally  shorter,  and  not  liable  to  such 
varying  loads,  as  is  the  case  with  bridges. 

Until  a  very  recent  date,  almost  the  only  experiments  on 
the  strength  of  cast-iron  columns  were  those  of  Hodgkinson, 
recorded  in  the  London  Philosophical  Transactions  for  1840. 

The  pillars  tested  were  mostly  of  cast-iron,  as  being  the 
material  in  most  general  use  at  that  time  for  that  purpose ; 
but  some  were  of  wrought-iron,  and  a  few  of  wood. 

Until  the  time  when  Hodgkinson  made  his  experiments,  the 
prevailing  theory  of  the  strength  of  columns  was  that  of  Euler, 
which  has  been  already  explained  in  §  210. 

According  to  this  theory,  a  column  of  any  given  dimensions 
and  material  requires  a  certain  weight  to  bend  it,  even  in  the 
slightest  degree ;  and,  with  less  than  this  weight,  it  would  not 
be  bent  at  all. 

This  weight  producing  incipient  flexure  is  to  be  found,  as 
we  have  already  seen,  by  the  formulae, 

P  =  (— -.)  El  for  one  end  fixed  in  direction  and  the  other 

\2//  ,     , 

rounded. 
P  =  ( ^}  El  for  both  ends  rounded. 

P  =  ( yj  El  for  both  ends  fixed  in  direction,  where  /  — 
length  of  column,  /  =  moment  of  inertia  of 
section,  E  =  modulus  of  elasticity  of  the  ma- 
terial. 

The  load  P  is,  then,  according  to  Euler,  the  breaking-load, 
unless  the  column  is  so  short  that  the  breaking-load  by  direct 
crushing  is  less  than  the  value  of  P,  in  which  case  the  column 
will  fail  by  direct  crushing.  Starting  from  this  point  of  view, 
Hodgkinson  says,  in  his  report,  "  My  first  object  was  to  supply 
the  deficiencies  of  Euler's  theory  of  the  strength  of  pillars  if  it 
should  appear  capable  of  being  rendered  practically  useful,  and, 
if  not,  to  endeavor  to  adapt  the  experiments  so  as  to  lead  to 


CAST-IRON  COLUMNS.  367 

useful  results."  He  also  says,  in  regard  to  Ruler's  theory,  "  I 
have  many  times  sought  experimentally,  with  great  care,  for  the 
weight  producing  incipient  flexure  according  to  the  theory  of 
Euler,  but  have  hitherto  been  unsuccessful.  So  far  as  I  can 
see,  flexure  commences  with  weights  far  below  those  with  which 
pillars  are  usually  loaded  in  practice.  It  seems  to  be  produced 
by  weights  much  smaller  than  are  sufficient  to  render  it  capable 
of  being  measured." 

Another  matter  to  which  Hodgkinson  devoted  considerable 
attention  in  this  investigation,  was  a  comparison  of  the  strength 
of  pillars  with  flat  and  with  rounded  ends  respectively.  By 
flat  ends  is  meant  having  the  ends  finished  in  planes  perpen- 
dicular to  the  axis,  and  having  the  resultant  of  the  weight  act 
along  the  axis  :  the  ends  are  then  fixed  in  direction.  By  rounded 
ends  Hodgkinson  actually  meant  rounded  ends ;  and  he  used 
such  in  his  experiments,  making  them  generally  hemispherical 
in  form.  The  results  have  been  assumed  generally  to  apply  to 
pin  ends,  or  to  any  irregularity  of  fixing  the  ends  which  does 
not  absolutely  fix  them  in  direction. 

As  a  result  of  his  experiments  he  states,  that,  in  all  pillars 
whose  length  is  thirty  times  the  diameter  or  upward,  the 
strength  of  those  with  flat  ends  seems  to  be  about  three  times 
as  great  as  the  strength  of  those  with  rounded  ends ;  the  mean 
ratio  being  3.167. 

In  pillars  shorter  than  thirty  diameters,  the  ratio  decreased 
when  the  ratio  of  the  length  to  the  diameter  decreased. 

The  experiments  were  all  made  on  circular  or  hollow  circular 
pillars,  the  lengths  varying  from  7-J  diameters  to  121  diameters  ; 
a  case  of  the  first  being  one  15!-  inches  long  and  0.51  inch  di- 
ameter, and  of  the  last,  30^  inches  long  and  i.oi  inch  diameter; 
the  largest  diameters  used  being  about  two  inches. 

The  empirical  formula  given  by  Hodgkinson  for  the  strength 
of  cast-iron  columns  more  than  thirty  diameters  long,  and  with 
flat  ends,  is 


368  APPLIED   MECHANICS. 


713-55  

^=99318--^-:  (i) 

where  D  =  outside  diameter  in  inches,  d  =  inside  diameter  in 
inches,  /  =  length  in  feet,  and  P  =  breaking-weight  in  pounds. 
For  columns  less  than  thirty  diameters  long,  Hodgkinson 
gives  the  following  formula  for  the  breaking-strength  :  — 

be  .  . 

y  =  TT-TJ  (2) 


where  c  =  force  which  would  crush  the  pillar  without  bending 
it,  b  =  breaking-weight  that  would  be  obtained  by  using  the 
formula  for  columns  more  than  thirty  diameters  long,  y  =  act- 
ual breaking-strength. 

For  determining  the  value  of  c,  we  have  as  the  crushing- 
strength  of  cast-iron,  from  some  experiments  of  Hodgkinson 
made  at  the  time  on  cast-iron,  the  value  109801  Ibs.  per  square 
inch  ;  so  that  we  should  have, 

For  solid  circular  columns, 


For  hollow  circular  columns, 

c  = 


For  columns  with  rounded  ends  more  than  thirty  diameters 
long,  the  breaking-weight  would  be  about  one-third  of  the 
result  for  flat  ends,  and,  by  assuming  it  as  one-third  for  shorter 
columns,  we  should  err  on  the  side  of  safety. 

Mr.  James  B.  Francis  has  computed  and  published  a  series 
of  tables  of  the  strength  of  cast-iron  pillars  of  the  ordinary  sizes 
used  in  practice,  the  computations  being  made  by  means  of 
Hodgkinson's  formulae  ;  the  safe  weights  given  in  his  tables 


CAST-IRON  COLUMNS.  369 

being  one-fifth  of  the  breaking-weights  of  rounded-end  columns, 
or  one-fifteenth  of  the  breaking-weights  as  computed  by  the 
formulae  for  flat  ends,  and  the  following  are  his  directions  in 
regard  to  the  use  of  his  tables  : 

"  For  practical  purposes  we  may  divide  the  use  of  cylin- 
drical cast-iron  pillars  into  three  cases : 

"  i°.  Pillars  with  rounded  ends,  or  put  up  as  they  come 
from  the  founder,  or  with  only  the  prominent  irregularities  at 
the  ends  chipped  off,  or  however  the  ends  may  be,  if  the  weight 
is  very  unequally  distributed  over  their  surfaces,  or  if  unequal 
settlements  may  be  apprehended.  When  the  designer  of  the 
pillars  has  no  control  over  the  execution  of  the  work,  or  any 
guaranty  that  his  directions  will  be  minutely  followed,  his  only 
safe  course  will  be  to  consider  them  in  this  case.  The  weights 
given  in  the  tables  are  the  safe  weights  the  pillars  will  support 
in  this  case. 

"  2°.  Pillars  with  the  ends  turned  in  a  lathe,  to  planes  at 
right  angles  to  the  axes,  and  put  up  with  ordinary  care,  so  that 
the  weight  will  be  distributed  over  the  whole  area  of  the  ends ; 
the  pillars  being  made  and  put  up  under  the  eye  of  the  person 
responsible  for  their  sufficiency;  or  with  such  other  guaranty 
for  faithful  workmanship  as  may  be  satisfactory  to  him.  In 
this  case,  it  is  assumed  that  the  safe  weight  is  one  and  a  half 
times  that  for  a  pillar  with  rounded  ends,  and  may  be  found 
from  the  tables  by  adding  one-half  to  the  tabular  weight. 

"  3°.  Pillars  with  the  ends  finished,  and  put  up  with  the 
degree  of  perfection  attained  in  Hodgkinson's  experiments  on 
pillars  with  flat  ends.  The  safe  weight  in  this  case  is  three 
times  that  given  in  the  tables." 

We  have  also  Gordon's  formulae,  which  Have  been  already 
given  in  §  209,  the  constants  of  which  are  based  upon  Hodg- 
kinson's experiments. 

Two  sets  of  tests  of  cast-iron  mill  columns  have  recently 
been  made  on  the  Government  testing-machine  at  Watertown 


37°  APPLIED  MECHANICS. 

Arsenal ;  an  account  of  these  sets  of  tests  is  published  in  their 
reports  of  1887  and  of  .1888. 

The  first  lot  consisted  of  eleven  old  cast-iron  columns,  which 
had  been  removed  from  the  Pacific  Mills  at  Lawrence,  Mass., 
during  repairs  and  alterations. 

The  second  lot  consisted  of  five  new  cast-iron  columns  cast 
along  with  a  lot  that  was  to  be  used  in  a  new  mill. 

Of  these  five,  the  strength  of  two  was  greater  than  the 
capacity  of  the  testing-machine,  hence  only  three  were  broken  ; 
while  in  the  case  of  the  other  two  the  test  was  discontinued 
when  a  load  of  800000  Ibs.  was  reached.  All  the  columns  con- 
tained a  good  deal  of  spongy  metal,  which  of  course  rendered 
their  strength  less  than  it  would  otherwise  have  been  ;  never- 
theless, inasmuch  as  this  is  just  what  is  met  with  in  building, 
it  is  believed  that  these  tests  furnish  reliable  information  as  to 
what  we  should  expect  in  practice,  and  that  this  information 
is  much  more  reliable  than  any  that  can  be  derived  from  test- 
ing small  columns. 

In  all  the  tests  the  compressions  were  measured  under  a 
large  number  of  loads  less  than  the  ultimate  strength ;  but  in- 
asmuch as  it  is  not  possible,  in  the  case  of  cast-iron,  to  fix  any 
limits  within  which  the  stress  is  proportional  to  the  strain,  no 
attempt  will  here  be  made  to  compute  the  modulus  of  elas- 
ticity. Hence  there  will  be  given  here  a  table  showing  the 
dimensions  of  the  columns  tested,  their  ultimate  strengths, 
and,  in  those  cases  where  they  were  measured,  the  horizontal 
and  vertical  components  of  their  deflections,  measured  at  the 
time  when  their  ultimate  strengths  were  reached,  as  the  Govern- 
ment machine  is  a  horizontal  machine.  A  glance  at  the  table 
will  make  it  evident  that  we  cannot,  in  the  case  of  such  columns, 
rely  upon  a  crushing  strength  any  greater  than  25000  or  30000 
Ibs.  per  square  inch  of  area  of  section.  Hence  it  would  seem 
to  the  writer  that,  in  order  to  proportion  a  cast-iron  column 
to  bear  a  certain  load  in  a  building,  we  should  first  compute 


CAST-IRON   COLUMNS. 


the  greatest  load  which  could  come  upon  the  column,  and,  di- 
viding this  by  5000,  which  would  correspond  to  a  factor  of 
safety  between  5  and  6,  obtain  an  area  of  section  which  might 
be  probably  adopted,  provided  it  proves  itself  sufficient  for  the 
other  requirements  which  are  to  follow  here.  Next,  determine 
the  outside  diameter  in  such  a  way  as  to  avoid  an  excessive 
ratio  of  length  to  diameter ;  if  this  ratio  is  not  much  in  excess 
of  twenty,  the  extra  stress  produced  by  any  eccentricity  of  the 
load  due  to  the  deflection  of  the  column  will  be  very  slight. 
At  the  same  time  see  that  the  thickness  of  metal  is  sufficient 
to  insure  a  good  sound  casting. 

Now,  having  figured  the  column  in  this  way,  compute  the 
outside  fibre  stress  (using  the  method  of  §  207)  that  would 
occur  with  the  loading  of  the  floors  assumed  to  be  such  as  to 
give  as  great  an  eccentricity  as  it  is  possible  to  bring  upon  the 
column.  If  this  distribution  of  the  load  is  one  that  is  likely 
to  occur,  then  the  maximum  fibre  stress  in  the  column  due  to 
it  ought  not  to  be  greatly  in  excess  of  5000  Ibs.  per  square 
inch ;  but  if  it  is  one  which,  there  is  scarcely  a  chance  of  realiz- 
ing, then  the  maximum  fibre  stress  under  it  might  be  allowed 
to  reach  IOOOO  Ibs.  per  square  inch.  If  by  adopting  the  di- 
mensions already  chosen  these  results  can  be  obtained,  we  may 
adopt  them  ;  but  if  it  is  necessary  to  increase  the  sectional 
area  in  order  to  accomplish  them,  we  should  increase  it. 

Another  matter  that  should  be  referred  to  here  is  the  fact 
that  a  long  cap  on  a  column  is  more  conducive  to  the  produc- 
tion of  an  eccentric  loading  than  a  short  one ;  hence,  that  a 
long  cap  is  a  source  of  weakness  in  a  column. 

Other  sources  of  weakness  in  cast-iron  columns  are  spongy 
places  in  the  casting  (which  correspond  in  a  certain  way  with 
knots  in  wood),  and  also  an  inequality  in  the  thickness  of  the 
two  sides  of  the  column,  the  result  of  this  being  the  same  as 
that  of  eccentric  loading ;  and  it  is  especially  liable  to  occur  in 
consequence  of  the  fact  that  it  is  the  common  practice  to  cast 


372 


APPLIED  MECHANICS. 


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APPLIED     MECHANICS. 


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374 


CAST-IRON    COLUMNS. 


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is 


CAST-IRON   COLUMNS. 


columns  on  their  side,  and  not  on  end.  The  engineer  should, 
however,  inspect  all  columns  to  be  used  in  a  building,  and  re- 
ject any  that  have  the  thickness  of  the  shell  differing  in  different 
parts  by  more  than  a  very  small  amount. 

Professor  Bauschinger,  of  the  Munich  laboratory,  made  two 
series  of  tests  of  full-size  cast-  and  of  wrought-iron  columns  in 
order  to  determine  the  effect  upon  their  strength  of  exposing 
them  to  the  heat  of  a  fire,  and  then  sprinkling  them  with  water 
from  a  hose  after  they  had  come  to  a  red  heat.  The  columns 
were  placed  in  his  (horizontal)  testing-machine,  and  loaded 
with  what  he  estimated  as  the  safe  load,  using  a  factor  of 
safety  five. 

The  formulae  which  he  used  in  the  first  series  of  tests  made 
in  1884,  and  reported  in  Heft  XII  of  the  Mittheilungen,  were, 
when  reduced  to  English  measures,  pounds  and  inches  being 
the  units  : 

For  cast-iron, 

19912.4 

/a ' 
i  -f-  0.0006-5 

For  wrought-iron, 

11378.4 

/" 
i  -f-  0.00009 ~j 

where  P  —  safe  load  (factor  of  safety  five),  A  =  area  of  section, 
/  =  length,  p  =  least  radius  of  gyration. 

A  fire  was  made  in  a  U-shaped  receptacle  under  the  post, 
so  arranged  that  the  flames  enveloped  the  post.  The  tem- 
perature was  determined  from  time  to  time  by  means  of  alloys 
of  different  melting-points  ;  and  the  horizontal  and  vertical 
components  of  the  deflections  were  read  off  on  a  dial  as  indi- 
cated by  a  hand  attached  to  the  post  by  a  long  wire.  The 
post  was  also  examined  for  cracks  or  fractures. 


APPLIED  MECHANICS. 


In  the  1884  series  he  tested  six  cast-iron  posts  of  various 
styles,  and  three  wrought-iron  posts,  one  of  them  being  made 
of  channel  irons  and  plates  put  together  with  screw-bolts,  one 
of  I  irons  and  plates  also  put  together  with  screw-bolts,  and 
one  hollow  circular.  He  measured  the  deflections  and  the 
temperatures  at  different  times,  and  while  he  tested  in  many 
cases  the  effect  of  sprinkling  at  lower  temperatures,  he  always 
tried  it  after  the  post  had  reached  redness,  or  else  about  600° 
Centigrade. 

He  generally  had  the  bolts  that  held  the  platforms  of  the 
machine  tight,  thus  fixing  the  ends  of  the  columns  in  direc- 
tion ;  but  in  some  cases  he  loosened  these  bolts,  and  thus  left 
the  ends  free  to  change  their  directions. 

He  says,  however,  that  this  latter  arrangement,  while  it 
weakens  the  post,  has  less  effect  than  the  manner  of  sprink- 
ling, i.e.,  the  part  of  the  post  wTiere  the  stream  is  thrown. 

The  details  of  the  tests  will  not  be  given  here,  but  only 
Bauschinger's  conclusions.  He  said  : 

That  wrought-iron  columns,  even  under  the  most  favorable 
adjustment  of  their  ends  and  of  the  manner  of  loading,  bend 
so  much  that  they  cannot  hold  their  load,  sometimes  with  a 
temperature  less  than  600°  Centigrade,  and  always  when  they 
are  at  a  red  heat  ;  and  this  bending  is  accelerated  by  sprink- 
ling on  the  opposite  side,  even  when  only  the  ends  of  the  post 
are  sprinkled. 

That  under  similar  circumstances  cast-iron  posts  bend,  and 
this  bending  is  increased  by  sprinkling  ;  but  it  does  not  exceed 
certain  limits,  even  when  the  post  is  red  for  its  entire  length 
and  the  stream  of  water  is  directed  against  the  middle,  and  the 
post  does  not  cease  to  bear  its  load  even^when  cracks  are  de- 
veloped by  the  sprinkling.  Only  when  both  ends  of  a  cast-iron 
post  are  free  to  change  their  directions  does  sprinkling  them 
at  the  middle  of  the  opposite  side  when  they  are  red  make 


CAST-IRON  COLUMNS.  377 

them  break,  but  such  an  unfavorable  case  of  fastening  the  ends 
hardly  ever  occurs  in  practice. 

That  the  cracks  in  the  columns  tested  occurred  in  the 
smooth  parts,  and  not  at  corners  or  projections. 

That  the  result  of  these  tests  warns  us  to  be  much  more 
prudent  in  regard  to  the  use  of  wrought-iron  in  building.  If 
posts  which  are  subjected  to  a  longitudinal  pressure  bend  so 
badly  when  subjected  to  heat  on  one  side  that  they  lose  the 
power  of  bearing  their  load,  how  much  more  must  this  be  the 
case  with  wrought-iron  beams  ;  and  he  urges  the  importance  of 
making  more  experiments. 

In  Heft  XV  of  the  Mittheilungen  he  says  that  the  results 
were  criticised  in  two  ways,  viz. : 

(a)  Moller,  in  an  article  published  in  the  Arch.  u.  Ing. 
Vereins  zu  Hamburg  of  October  2ist,  1885,  claimed  that 
Bauschinger  had  not  used  correct  constants  in  the  formulae 
by  which  he-computed  the  safe  loads,  and  hence  that  the  loads 
he  had  put  on  the  cast-iron  columns  were  very  light  in  com- 
parison with  those  which  he  put  on  the  wrought-iron. 

(V)  Gerber  claimed  that  the  built-up  wrought-iron  posts 
which  Bauschinger  used  were  not  properly  made,  and  that  they 
should  have  had  more  bolts  or  rivets  and  been  better  stiffened 
sideways. 

Bauschinger  therefore  concluded  to  make  a  new  set  of  tests, 
and  for  this  purpose  he  had  made  two  cast-iron  and  five 
wrought-iron  columns — the  former  being  carefully  cast,  but  on 
the  side,  while  the  wrought-iron  ones  were  made  by  a  bridge 
company  of  very  good  reputation,  and  four  of  them  were  similar 
to  those  made  at  the  time  for  a  new  warehouse  in  Hamburg. 

Two  of  the  five  were  made  of  channels  and  plates  riveted 
together  throughout  their  entire  length,  making  a  box  section. 
The  other  three  were  each  made  of  four  angle-irons  set  in  the 
form  of  a  cross,  with  spaces  between  them,  but  fastened  to- 
gether by  other  irons  riveted  to  them  at  intervals.  The  first 


APPLIED  MECHANICS. 


two  continued  to  bear  their  load  after  heating  and  sprinkling, 
but  not  the  last  three ;  while  the  cast-iron  columns  behaved  in 
a  similar  way  to  what  they  did  in  the  previous  set  of  tests. 

He  now  used  for  the  wrought-iron  posts  the  same  formula 
as  before,  since  it  gives  lower  values  than  Gerber's  formula, 

14223.4 

~          ooooi-' 
°V 

For  the  two  cast-iron  posts,  on  the  other  hand,  he  now 
used  the  following  formula,  which  gave  very  considerably 
higher  results  than  that  which  he  used  in  the  other  tests: 

p__        19912.4 
i  +  0.00025^ 

The  tests  were  made  just  as  before,  and  the  following  are 
his  conclusions : 

That  when  wrought-iron  posts  are  as  well  constructed  as 
the  two  referred  to,  they  resist  fire  and  sprinkling  tolerably 
well,  though  not  as  well  as  cast-iron ;  but  that  posts  con- 
structed like  the  other  three,  even  with  the  fire  alone,  and 
before  the  sprinkling  begins,  get  so  bent  that  they  can  no 
longer  hold  their  load.  Good  construction  requires  that  the 
rows  of  rivets  shall  extend  through  the  entire  length  of  the 
post,  and  the  rivets  should  be  quite  near  each  other ;  but  the 
tests  are  not  extensive  enough  to  show  what  are  the  necessary 
requirements  to  make  wrought-iron  posts  able  to  stand  fire  and 
sprinkling;  in  order  to  know  this  more  experiments  are  needed. 

In  Dingler's  Polytechnisches  Journal  for  1889,  page  259  et 
seq.,  is  an  article  by  A.  Martens,  an  engineer  in  Berlin,  upon 
the  behavior  of  cast-  and  wrought-iron  in  fires,  considering  es- 
pecially the  burning  of  a  large  warehouse  in  Berlin,  and  advo- 
cating the  protection  of  iron  work  by  covering  it  with  cement. 


COLUMNS.        '  379 


He  says  that  there  are  two  series  of  tests  upon  this  subject, 
one  of  which  is  the  tests  of  Bauschinger  already  explained,  and 
the  other  a  set  of  tests  made  by  Moller  and  Luhmann. 

The  following  account  of  the  latter  is  condensed  from  Mr. 
Martens's  paper.  The  account  was  published  in  the  Verhand- 
lung  der  Verein  fur  Gewerbefleis  for  1887,  page  573  et  seq. 

The  tests  were  made  in  a  much  cruder  manner  than  those 
of  Bauschinger,  as  the  machine  used  was  a  hydraulic  press,  the 
load  being  weighed  by  a  gauge  on  the  cylinder,  hence  render- 
ing the  results  subject  to  the  error  of  the  piston  friction  —  an 
unknown  and  sometimes  a  very  considerable  quantity  ;  also, 
the  deflections  were  measured  by  a  lever. 

He  says  that  Moller  claims  that  posts  in  use  areliot  loaded 
centrally,  but  that  the  load  is  generally  eccentric,  and  that  he 
loaded  all  of  his  posts  in  such  a  way  that  the  resultant  pressure 
was  applied  about  0^.4  out  of  centre,  and  that  Moller  gives  a 
formula  which,  in  accordance  with  his  experiments,  will  give 
the  load  a  post  can  safely  bear  under  such  circumstances  when 
subjected  to  fire  and  sprinkling. 

The  tests  were  made  on  posts,  the  area  of  whose  section 
was  about  9.6  or  9.9  sq.  in.,  and  of  lengths  of  1,2,  and  4  metres 
respectively  for  the  cast-iron  and  of  I  and  2  metres  only  for 
the  wrought-iron.  The  hollow  ones  had  an  outside  diameter 
of  about  5".9  and  thickness  of  about  o".6.  The  solid  ones  had 
a  diameter  of  about  3".  5.  Also,  riveted  posts,  of  the  same  area 
of  section  and  2  metres  long,  were  tested. 

Besides  this  he  tested  a  cast-  and  wrought-iron  post,  each 
protected  by  a  coating  of  cement  about  2".  5  thick;  a  riveted 
post  covered  with  pine  about  o"  .9  thick;  and  a  cast-  and  a 
wrought-iron  post,  each  with  a  2"  gas-pipe  in  the  middle,  with 
the  space  between  filled  with  cement. 

Moller's  conclusions,  as  stated  by  Martens,  are  as  follows  : 

i°.  With  ten  cast-iron  posts  he  could  not  get  any  cracks 
by  sprinkling  at  a  red  heat  ;  but  it  is  to  be  noted  that  his  were 


3^0  APPLIED   MECHANICS. 


new  posts,  while  those  used  in  Bauschinger's  first  series  were 
old  ones,  and  that  those  in  Bauschinger's  second  series,  which 
were  new  and  very  carefully  cast,  did  not  show  cracks  either. 

2°.  He  claims  that  while  the  cracks  would  allow  the  post 
still  to  bear  a  centre  load,  it  could  not  bear  an  eccentric  load 
or  a  transverse  load. 

3°.  He  claims  that  the  load  on  a  cast-iron  post  should  be 
limited  to  one  which  shall  not  produce  sufficient  bending  to 
bring  about  a  tensile  stress  anywhere  when  the  post  is  bent  by 
the  heat  and  sprinkling. 

4°.  He  claims  that  in  either  cast-  or  wrought-iron  posts,  if 
the  ends  are  not  fixed,  the  ratio  of  length  to  diameter  should 
not  exceed  10,  whereas  if  they  are  it  should  not  exceed  17; 
also,  that  there  is  no  such  thing  as  absolute  safety  from  fire 
with  iron. 

5°.  A  covering  of  cement  delays  the  action  of  the  fire,  and 
that  therefore  such  a  covering  is  a  protection  to  the  post 
against  excessive  one-sided  heating  and  cooling. 

6°.  Cast-iron  is  more  likely  to  have  at  any  one  section  a 
collection  of  hidden  flaws  than  wrought-iron. 

By  way  of  comment  I  should  say : 

i°.  The  inaccuracy  of  his  testing-machine,  which  may  have 
been  wrong  by  10  %,  or  even  more,  necessarily  vitiates  any  for- 
mula he  might  devise  for  the  cases  he  tried,  viz.  :  o".^  eccen- 
tricity. 

2°.  The  eccentricity  o".4  does  not  represent  what  occurs  in 
practice,  as  the  latter  are  variable,  and  usually  far  larger  than 

o".4. 

3°.  The  fact  that  Bauschinger  found  cracks  when  testing 
old  posts,  whether  they  be  due  to  eccentricity,  overstraining, 
or  originally  poor  quality,  proves  that  cracks  are  likely  to  occur. 

The  conclusions  to  which  I  have  not  referred  in  these  com- 
ments are  not  based  upon  the  experiments,  but  upon  Holler's 
judgment. 


TRANSVERSE   STRENGTH  OF  CAST-IRON.  381 

§  220.  Transverse  Strength  of  Cast-Iron. — At  one  time 
cast-iron  was  very  largely  used  for  beams  and  girders  in  build- 
ings to  support  a  transverse  load.  Its  use  for  this  purpose  has 
now  been  almost  entirely  abandoned,  as  it  has  been  superseded 
by  wrought-iron  and  steel. 

A  great  many  experiments  have  been  made  on  the  trans- 
verse strength  of  cast-iron ;  the  specimens  used  in  some  cases 
being  small,  and  in  others  large.  The  records  of  a  great  many 
experiments  of  this  kind  are  to  be  found  in  the  first  four  books 
of  the  list  already  enumerated  in  §  218.  The  details  of  these 
tests  will  not  be  considered  here,  but  an  outline  will  be  given 
of  some  of  the  main  difficulties  that  arise  in  applying  the  results 
and  in  using  the  beams. 

Cast-iron  is  treacherous  and  liable  to  hidden  flaws ;  it  is 
brittle.  It  is  also  a  fact  that  in  casting  any  piece  where  the 
thickness  varies  in  different  parts,  the  unequal  cooling  is  liable 
to  establish  initial  strains  in  the  metal,  and  that  therefore 
those  parts  where  such  strains  have  been  established  have 
their  breaking-strength  diminished  in  proportion  to  the  amount 
of  these  strains. 

In  the  case  of  cast-iron  also,  the  ratio  of  the  stress  to  the 
strain  is  not  constant,  even  with  small  loads,  and  is  far  from 
constant  with  larger  loads  ;  also,  inasmuch  as  the  compressive 
strength  is  far  greater  than  the  tensile,  it  follows  that,  in  a 
transversely  loaded  beam  which  is  symmetrical  above  and  be- 
low the  middle,  the  fibres  subjected  to  tension  approach  their* 
full  tensile  strength  long  before  those  subjected  to  compression 
are  anywhere  near  their  compressive  strength.  The  result  of 
all  this  is,  that  if  a  cast-iron  beam  be  broken  transversely,  and 
the  modulus  of  rupture  be  computed  by  using  the  ordinary 
formula, 

,_  My 
f~~T> 


APPLIED  MECHANICS. 


we  shall  find,  as  a  rule,  a  very  considerable  disagreement  be- 
tween the  modulus  of  rupture  so  calculated  and  either  the 
tensile  or  compressive  strength  of  the  same  iron.  Indeed, 
Rankine  used  to  give,  as  the  modulus  of  rupture  for  rectangu- 
lar cast-iron  beams,  40000  Ibs.  per  square  inch,  and  for  open- 
work beams  17000  Ibs.  per  square  inch,  which  latter  is  about 
the  tensile  strength  of  fairly  good  common  cast-iron. 

A  great  deal  has  been  said  and  written,  and  a  good  many 
experiments  have  been  made,  to  explain  this  seeming  disagree- 
ment between  the  modulus  of  rupture  as  thus  computed,  and 
the  tensile  strength  of  the  iron.  Barlow  proposed  a  theory 
based  upon  the  assumption  of  the  existence  of  certain  stresses 
in  addition  to  those  taken  account  of  in  the  ordinary  theory  of 
beams,  but  his  theory  has  no  evidence  in  its  favor. 

Rankine  claimed  that  the  fact  that  the  outer  skin  is  harder 
than  the  rest  of  the  metal  would  serve  to  explain  matters,  but 
this  would  not  explain  the  fact  that  the  discrepancy  exists  in 
the  case  of  planed  specimens  also. 

Neither  Barlow  nor  Rankine  seems  to  have  attempted  to 
find  the  explanation  in  the  fact  that  the  formula 

My 


assumes  the  proportionality  of  the  stress  to  the  strain,  and 
hence  that  is  less  and  less  applicable  the  greater  the  load,  and 
hence  the  nearer  the  load  is  to  the  breaking  load.  An  article 
by  Mr.  Sondericker  in  the  Technology  Quarterly  of  October, 
1888,  gives  an  account  of  some  experiments  made  by  him  to 
test  the  theory  that  "  the  direct  stress,  tension,  or  compression, 
at  any  point  of  a  given  cross-section  of  a  beam,  is  the  same 
function  of  the  accompanying  strain,  as  in  the  case  of  the  cor- 
responding stress  when  uniformly  distributed,"  and  the  results 
bear  out  the  theory  very  well  ;  hence  it  follows  that,  if  we  use 
the  common  theory  of  beams,  determining  the  stresses  as  such 


TRANSVERSE   STRENGTH  OF  CAST-IRON.  383 

multiples  of  the  strains  as  they  show  themselves  to  be  in  direct 
tensile  and  compressive  tests,  the  discrepancies  largely  vanish, 
and  those  that  are  left  can  probably  be  accounted  for  by  initial 
stresses  due  to  unequal  rate  of  cooling,  and  by  the  skin,  or 
by  lack  of  homogeneity.  In  the  same  article  he  quotes  the 
results  of  other  tests  bearing  more  or  less  on  the  matter,  and 
there  will  be  quoted  here  the  table  on  the  next  page. 
If,  therefore,  we  wish  to  make  use  of  the  formula 

M=f~ 

y 

in  calculating  the  strength  of  cast-iron  beams,  we  cannot  use 
one  fixed  value  of  f  for  all  beams  made  of  one  given  quality 
of  cast-iron,  but  we  shall  have  to  use  a  very  varying  modulus 
of  rupture,  varying  especially  with  the  form,  and  also  with  the 
size  of  the  beam  under  consideration.  Now,  in  order  to  do 
this,  and  obtain  reasonably  correct  results,  we  need,  wherever 
possible,  to  use  values  of  f  that  have  been  deduced  from  ex- 
periments upon  pieces  like  those  which  we  are  to  use  in  prac- 
tice, and  under,  as  nearly  as  possible,  like  conditions. 

There  are  not  very  many  records  of  such  experiments  avail- 
able, and,  in  cases  where  we  cannot  obtain  them,  it  will  prob- 
ably be  best  to  use  a  value  of  f  no  greatei  than  the  tensile 
strength  for  complicated  forms,  and  forms  having  thin  webs. 
For  pieces  of  rectangular  or  circular  section  we  might  probably 
use,  for  good  fair  cast-iron,  25000  to  30000  Ibs.  per  square 
inch. 

A  few  tests  of  the  character  referred  to  have  been  made  in 
the  engineering  laboratories  of  the  Massachusetts  Institute  of 
Technology,  and  a  brief  statement  of  them  will  be  given  here. 
The  first  that  will  be  referred  to  here  is  a  series  of  experiments 
made  by  two  students  of  the  Institute,  an  account  of  which  is 
given  in  the  Proceedings  of  the  American  Society  of  Mechani- 
cal Engineers  for  1889,  pp.  187  et  seq. 

The  object  of  this  investigation  was  to  determine  the  trans- 


APPLIED   MECHANICS. 


Modulus  of 

Form  of 
Beam    Sec- 

Tensile 
Strength, 

Rupture 

,      My 
J  —  ~.    i 

Ratio. 

Condition 
of 

Experimenter. 

tion. 

Ibs.  per  Sq. 

1 

Specimen. 

In. 

Ibs.  per  Sq. 

In. 

19850 

41320 

2.08 

Turned 

C.  Bach.* 

X^^x 

16070 

35500 

2.21 

Turned 

Considere.  f 

34420 

63330 

1.84 

Turned 

Considere. 

^n^ 

24770 

54390 

2.19 

Turned 

Robinson  and  Segundo.  \ 

25040 

46280 

1.85 

Rough 

Robinson  and  Segundo. 

Mean. 

2.03 

16070 

29250 

1.82 

Planed 

Considere. 

ilfp 

36270 

58760 

1.62 

Planed 

Considere. 

19090 

33740 

1-77 

Planed 

C.  Bach. 

Mean. 

1.74 

19470 

34000 

1-75 

Planed 

C.  Bach. 

Hi 

31430 

49030 

1.56 

Planed 

Considere. 

m 

19880 

33860 

1.70 

Planed 

Sondericker. 

24770 

42340 

1.71 

Planed 

Robinson  and  Segundo. 

P^I 

25040 

42IIO 

1.68 

Rough 

Robinson  and  Segundo. 

Mean. 

1.6.8 

19470 

28150 

1-45 

Planed 

C.  Bach. 

16070 

22500 

1.40 

Planed 

Considere. 

: 

31860 

36640 

i.i5 

Planed 

Considere. 

25040 

3I3IO 

1.25 

Rough 

Robinson  and  Segundo. 

Mean. 

I-3I 

16070 

23780 

1.48 

Planed 

Considere. 

H 

31290 

34730 

I.  II 

Planed 

Considere. 

H 

18050 

24550 

1.36 

Planed 

Sondericker. 

^^ 

22470 

26150 

1.16 

Rough 

Burgess  and  Viele.  § 

Mean. 

1.28 

*See  Zeitschrift  des  Vereines  Deutscher  Ingenieure,  Mar.  3d  and  loth, 

t  See  Annales  des  Fonts  et  Chausse"es.  1885. 

$See  Proceedings  Institute  of  Civil  Engineers,  Vol  86. 

§  See  Proceedings  Am.  Soc.  Mechl.  Engrs.  1889,  pp.  187  et  seq. 


TRANSVERSE  STRENGTH  OF  CAST-IRON. 


385 


verse  strength  of  cast-iron  in  the  form  of  window  lintels,  and 
also  the  deflections  under  moderate  loads,  and  from  the  latter 
to  deduce  the  modulus  of  elasticity  of  the  cast-iron,  and  to 
compare  it  with  the  modulus  of  elasticity  of  the  same  iron,  as 
determined  from  tensile  experiments  ;  also  the  tensile  strength 
and  limit  of  elasticity  of  specimens  taken  from  different  parts 
of  the  lintel  were  determined. 

The  iron  used  was  of  two  qualities,  marked  P  and  5  respec- 
tively. 

The  tensile  specimens  were  cast  at  the  same  time,  and  from 
the  same  run  as  the  lintels. 

Besides  this,  one  of  each  kind  of  window  lintels  was  cut  up 
into  tensile  specimens,  and  the  specimens  were  so  marked  as  to 
show  from  what  part  of  the  lintel  they  were  cut. 

The  tables  of  tests  will  now  be  given,  and  the  following  ex- 
planation of  the  symbolism  employed. 

P  and  5  are  used,  as  already  stated,  to  denote  the  quality 
of  the  iron. 

A  and  B  are  used  to  denote,  respectively,  that  the  specimen 
was  unplaned  or  planed. 

I,  2,  3,  etc.,  denote  the  number  of  the  test  made  on  that 
particular  kind  and  condition. 


I.,  II.,  III.,  denote  that  the  piece  has  been  taken  from  a 
lintel,  and  also  from  what  part,  as  will  easily  be  seen  by  the  ac- 
companying s'ketch. 


386 


APPLIED   MECHANICS. 


Thus  P.  B.  3  would  signify  that  the  specimen  was  of  quality 
P,  had  been  planed,  and  was  the  third  test  of  this  class. 

On  the  other  hand,  P.  B.  3  II.,  would  signify  in  addition 
that  it  had  been  taken  from  a  lintel,  and  was  a  piece  of  one  of 
the  strips  marked  II.  in  the  sketch. 

The  following  is  a  summary  of  the  breaking-weights  per 
square  inch  of  the  specimens  not  cut  from  the  lintels : 


p-  A-  i 23757 

P.  A.  2 21423 

P.  A.3 18938 

P.  A.  4 21409 

4)85527 


S.  A.  i 24204 

S.  A.  2... 25258 

S.  A.  3 24706 


P.  B.  i 
P.  B.  3 


21382 

21756 

25207 


3)74168 


24723 


S.  B.  i 29574 

S.  B,  2 23201 


2)46963 


2)52775 


23482  26388 

The  following  are  the  breaking-weights  per  square  inch  of 
the  specimens  cut  from  the  lintels  : — 


P.  B. 


'  5 

I 

6 

I 

9 

I 

10 

I 

4 

II 

7 

II 

ii 

II 

12 

II 

2 

III 

13 

III 

I  8 

IV 

19651 
20715 
21076 

21483 
19016 

19376 
22146 
20552 

"594 
16141 
19616 


S.  B. 


6 

7 
8 

3 

4 

5 

9 

10 

4 


I 
I 
I 

II 

II 

II 

III 

III 

IV 


29124 
28372 

25425 
24704 
29414 
23610 

27523 
18301 
19616 


TRANSVERSE  STRENGTH  OF  CAST-IKON. 


387 


All  the  window  lintels  tested  were  of  the  form  shown  in 
the  figure,  and  all  were  supported  at  the  ends  and  loaded  at 
the  middle,  the  span  in  every  case  being  52".  From  the  cut 
it  will  be  seen  that  the  web  varied  in  height,  being  4  inches 
high  above  the  flange  in  the  centre,  and  decreasing  to  2.5  inches 
at  the  ends  over  the  supports. 

The  following  are  the  results  of  the  separate  tests,  where 
tensile  modulus  of  rupture  means  the  outside  fibre  stress  per 
square  inch  on  the  tension  side,  and  compressive  modulus  of 
rupture  that  on  the  compression  side,  both  being  calculated 
from  the  actual  breaking  load  by  the  formula 


Mark  on  Lintel. 

Breaking-Load, 
Lbs 

Tensile  Modulus  of 
Rupture, 

Compressive  Modulus 
of  Rupture, 

Ibs.  per  Sq.  In. 

Ibs.  per  Sq.  In. 

P.    I 

27220 

26648 

81578 

P.    2 

30520 

29879 

91467 

P.  3 

27200 

26659 

81608 

S.    i 

26750 

26198 

80164 

S.     2 

19850 

J9433 

59490 

S-   3 

28670 

28068 

85924 

S.   4 

25120 

24592 

75285 

The  second  series  of  experiments  was  made  by  two  other 
students,  and  an  account  of  the  work  is  given  in  the  same 
article  as  the  former  one. 

The  object  was  to  determine  the  constants  suitable  to  use 
in  the  formulae  for  determining  the  strength  of  the  arms  of 
cast-iron  pulleys  ;  and  also,  incidentally,  to  determine  the  hold- 
ing power  of  keys  and  set-screws. 

Some  old  pulleys  with  curved  arms,  which  had  been  in  use 
at  the  shops,  were  employed  for  these  tests.  They  were  all 


$38  APPLIED   MECHANICS. 

about  fifteen  inches  in  diameter,  and  were  bored  for  a  shaft 
IT\  inches  in  diameter. 

Inasmuch  as  this  size  of  shaft  would  not  bear  the  strain 
necessary  to  break  the  arms,  the  hubs  were  bored  out  to  a 
diameter  of  \\\  inches  diameter,  and  key-seated  for  a  key  one- 
half  an  inch  square. 

In  order  to  strengthen  the  hubs  sufficiently,  two  wrought- 
iron  rings  were  shrunk  on  them,  so  as  to  make  it  a  test  of  the 
arms  and  not  of  the  hub. 

The  pulley  under  test  is  keyed  to  a  shaft  which,  in  its  turn, 
is  keyed  to  a  pair  of  castings  supported  by  two  wrought-iron  I- 
beams,  resting  upon  a  pair  of  jack-screws,  by  means  of  which 
the  load  is  applied.  A  wire  rope  is  wound  around  the  rim  of 
the  pulley,  and  leaves  it  in  a  tangential  direction  vertically. 
This  rope  is  connected  with  the  weighing  lever  of  the  machine, 
and  weighs  the  load  applied. 

In  a  number  of  the  experiments  one  arm  gave  way  first, 
and  then  the  unsupported  part  of  the  rim  broke. 

The  breaking-load  of  the  separate  pulleys  was,  of  course, 
determined,  and  then  it  was  sought  to  compute  from  this  the 
value  of  f  from  the  formula 


which  is  the  one  most  commonly  given  for  the  strength  of 
pulley  arms,  and  which  is  based  upon  several  erroneous  assump- 
tions, one  of  which  is  that  the  bending-moment  is  equally 
divided  among  the  several  arms.  In  this  formula 

/=  moment  of  inertia  of  section, 

n  =  number  of  arms, 

y  =  half  depth  of  each  arm  =  distance  from  neutral  axis  to 
outside  fibre, 

x  =  length  of  each  arm  in  a  radial  direction, 

P  =  breaking-load  determined  by  experiment. 


TRANSVERSE   STRENGTH  OF  CAST-IRON. 


389 


The  results  are  given  in  the  following  table,  the  units  being 
inches  and  pounds : 


g  g 

rt    rt 

<u    v 

6  6 
'S'S 

Si 

a    rt 

II 


}! 

t! 


<= 


J     x  o 

£|  S" 
|a  il 

II  I! 


il 
o"" 


TO  * 

a  cu'S 

CO  J3    &       . 

«  *^  fv 


£     «5.a.M 
2     <G  -  o 


b£ 

c 

a    E 


5  J3 


i  i 


II  v 

U     O      J3' 

ic  5  ^ 


•°  a      a 

)  3      v^  ^ 


c  a 
O 


V  3 

O 


I 


OO          CO         - 


.2    « 


XXX         X 


•B 

X 


It^X 

X 


X 

«p 


io|oo 

X 


X     X 

+  * 


«p*-p 


XXX        X        X        X        X 


* 

X 


i*r 

X 


X      X 


•sauv  jo  jaqain^ 


VO 


vO 


stn.iv 


•<t       co 


•qnn  jo  ss3U3piqx 


^£     ^     »JS 


HD<>0|tCr4*  1-fH 

rJ-rn|rH  CO  CO 

CO 


•mm  jo  ssaujpiqx 


-P  -P  -P  •* 


rh  CO  CO        CO 


jo  - 


* 


•1S3X  jo  J3qran>i 


oo        ON       O        « 


390 


APPLIED  MECHANICS. 


In  the  cases  of  numbers  5,  7,  8,  9,,  and  10  some  of  the  arms 
were  not  broken,  the  rims  were  now  broken  off,  and  the  re- 
maining arms  were  tested  separately,  the  pull  being  exerted  by 
a  yoke  hung  over  the  end  of  the  arm,  the  lower  end  being  at- 
tached to  the  link  of  the  machine. 

The,  arms  were  always  placed  so  that  the  direction  of  the 
pull  was  tangent  to  the  curve  of  the  rim  at  the  end  of  the  arm. 
The  actual  outside  fibre  stress  at  fracture  was  then  determined 
by  calculation  from  the  experimental  results,  and  is  recorded 
in  the  following  table,  the  units  being  inches  and  pounds  : — 


Number  of 
Arms. 

Dimensions  of  Sec- 
tion at  Fracture  : 
all  elliptical. 

Bend   of  Arm  with 
or  against  Load. 

Actual  Outside 
Fibre     Stress     at 
Fracture. 

AverageModulus  of 
Rupture  for  each 
Pulley. 

5  —  i 

TT9*    X    -JJ 

against 

45396 

45396 

7  —  i 

ii     X  } 

against 

36802 

7  —  2 

ijf    X    £ 

against 

39537 

7  —  3 

itt.x.i 

with 

46407 

40915 

8  —  i 

lit  x  it 

against 

35503 

8  —  2 

iii  X  If 

against 

36091 

8-3 

iii  x  it 

with 

39939 

8-4 

itt  x  it 

with 

42469 

3850° 

9  —  1 

iA  x  f 

against 

41899 

9  —  2 

IT*  X  H 

against 

44148 

9  —  3 

iA  X  f 

with 

55442 

47163 

10  —  i 

if     X  it 

against 

54743 

IO  2 

iff  X  it 

against 

5°943 

10  —  3 

lit  X  it 

against 

38605 

10  —  4 

ij   x  it 

with 

55229 

49880 

Total 663153 

Average 44210 


IV R  0  UGHT-IR  ON.  3  9 1 


§221.  Wfought-Iron.  —  Wrought-iron  is  the  product  ob- 
tained by  removing  the  carbon  from  cast-iron.  It  is  produced 
by  melting  the  iron,  and  passing  an  oxidizing  flame  over  it. 
When  the  carbon  is  burned  out,  the  mass  of  iron  is  left  in  a 
pasty  condition,  full  of  holes.  It  is  then  taken  out,  and  ham- 
mered or  rolled  in  order  to  unite  it  into  one  mass.  Wrought- 
iron  is  thus,  from  the  commencement  of  its  manufacture,  a 
series  of  welds ;  and  the  perfection  or  imperfection  of  these 
welds  affects  very  seriously  the  quality  of  the  iron. 

The  result  of  this  first  process  is  not  suitable  to  use  in  any 
construction  of  importance  ;  but  it  requires  to  be  re-heated 
and  re-rolled  a  number  of  times,  in  order  to  make  it  more  homo- 
geneous, and  to  remove  flaws  from  within  the  iron. 

At  best,  however,  wrought-iron  is  a  series  of  welds  ;  and,  if 
a  piece  be  broken,  the  separate  layers  of  which  it  is  composed 
can  be  seen  plainly.  It  is  also  subject  to  the  impurities  of  the 
cast-iron  from  which  it  is  made.  Thus,  the  presence  of  sulphur 
makes  it  red-short,  or  brittle  when  hot ;  and  the  presence  of 
phosphorus  makes  it  cold-short,  or  brittle  when  cold. 

It  cannot,  like  cast-iron,  be  melted  and  run  into  moulds : 
but  it  can  be  easily  welded  by  forging :  that  is,  two  masses  of 
wrought-iron  can  be  united  by  being  brought  to  a  proper 
temperature,  and  then  hammered  together. 

Wrought-iron  is  much  more  capable  of  bearing  a  tensile  or 
transverse  stress  than  cast-iron :  it  is  tougher,  it  stretches 
more,  and  gives  more  warning  before  fracture.  At  one  time 
cast-iron  was  almost  the  only  form  in  which  iron  was  used 
in  construction ;  but  now  wrought-iron  and  steel  are  super- 
seding it  in  by  far  the  majority  of  cases  where  strength 
and  toughness,  and  the  ability  to  resist  various  stresses,  are 
demanded. 

Wrought-iron  is  also  expected  to  withstand  a  great  many 
trials  that  would  seriously  injure  cast-iron :  thus,  two  pieces 
of  wrought-iron  are  generally  united  together  by  riveting;  the 


392  APPLIED  MECHANICS. 

holes  for  the  rivets  have  to  be  punched  or  drilled,  and  then 
the  rivets  have  to  be  hammered  ;  the  entire  process  tending  to 
injure  the  iron.  Wrought-iron  has  to  withstand  flanging,  and  is 
liable  to  severe  shocks  when  in  use ;  as,  for  instance,  those  that 
occur  from  the  difference  of  temperature,  and  the  changes  of 
temperature  in  the  different  parts  of  a  steam-boiler. 

§  222.  Tensile  and  Compressive  Strength  of  Wrought- 
Iron.  —  As  to  the  experimenters  on  the  tensile  strength  and 
elasticity  of  wrought-iron,  those  who  preceded  Hodgkinson  are 
of  little  more  than  historic  interest.  The  following  list  includes 
a  number  of  the  most  known  experimenters  :  — 

i°.  Eaton  Hodgkinson  :   (a)  Report  of  Commissioners  on  the  Applica- 
tion of  Iron  to  Railway  Structures. 
(b)  London  Philosophical  Transactions.     1840. 
2°.  William  H.  Barlow  :  Barlow's  Strength  of  Materials. 
3°.  Sir  William   Fairbairn  :    On  the  Application  of  Cast  and  Wrought 

Iron  to  Building  Purposes. 
4°.  Franklin  Institute  Committee  :    Report  of   the    Committee  of  the 

Franklin  Institute.     In  the  Franklin  Institute  Journal  of  1837. 
5°.  L.  A.  Beardslee,  Commander,  U.S.N. :  Experiments  on  the  Strength 

of  Wrought-iron  and  of  Chain  Cables.     Revised  and  enlarged  by 

William  Kent,  M.E.,  or  Executive  Document  98,  451)1  Congress, 

as  stated  below. 

6°.  David  Kirkaldy  :  Experiments  on  Wrought-iron  and  Steel. 
7°.  Professor  Bauschinger  :  Mittheilungen  aus  dem  Mech.-Tech.  Labora- 

torium  der  K.  Pol.  Schule  in  Mlinchen. 
8°.  G.  Bouscaren  :    Report  on  the  Progress  of  Work  on  the  Cincinnati 

Southern  Railway,  by  Thomas  D.  Lovett.     Nov.  i,  1875, 
9°.      Government  Testing  Machine  at  Watertown,  and  Government 
Commission :    (a)  Executive    Document    98,  45th   Congress 
U.S.A.,  2d  session. 
(d)   Exec.  Doc.  23,  46th  Congress  U.S.A.,  2d  session,  House. 

(c)    Exec.  Doc.  12,  47th  Congress  U.S.A.,  ist  session,  House. 

(d}  Exec.  Doc.  i,  47th  Congress  U.S.A.,  2d  session,  Senate. 

(<?)    Exec.  Doc.  5,  48th  Congress  U.S.A.,  ist  session,  Senate. 


TENSILE,  ETC.,  S TRENG TH  OF  WRO UGHT-IRON.       3 93 


(/)  Exec.  Doc.  35,  49th  Congress  U.S.A.,  ist  session,  Senate. 
(g)  Exec.  Doc.  36,  49th  Congress  U.S.A.,  ist  session,  Senate. 
(h)  Exec.  Doc.  31,  49th  Congress  U.S.A.,  2d  session,  House, 

Parts  I.  and  II. 
(k)  Tests  of  Metals  made  at  Watertown  Arsenal,  Mass.,  U.S.A., 

June  1887. 
10°.  A.    Wohler :    (a)    Die    Festigkeits   versuche    mit    Eisen    und 

Stahl. 

(b)  Strength  and  Determination  of  the  Dimensions  of  Struc- 
tures of  Iron  and  Steel,  by  Dr.  Phil.  Jacob  J.  Weyrauch. 
Translated  by  Professor  Dubois. 
11°.  Alexander   Holley  :    Executive  Document  23,  46th   Congress 

U.S.A.,  2d  session. 
12°.  Professor  R.  H.  Thurston  :  Materials  of  Engineering. 

A  few  tests  were  made  on  the  tensile  strength  of  wrought- 
iron  by  Eaton  Hodgkinson :  two  of  these  were  on  the  tensile 
strength  and  elasticity  of  rods  about  fifty  feet  long  ;  each  rod  be- 
ing made  in  three  parts,  these  parts  being  united  by  couplings. 

Below  is  given  the  table  of  results  of  the  first  of  these  tests, 
to  which  is  added  here  the  column  of  ratio  of  stress  per  square 
inch  to  total  strain : 


Weights 
applied,  in 
Ibs. 

Exten- 
sion, in 
inches. 

Sets,  in 
inches. 

Weights 
per  Square 
Inch  of 
Section,  in 
Ibs. 

Strains,  in 
Fractions 
of  the 
Length. 

Sets,  in  Frac- 
tions of 
the  Length. 

Ratio  of  Stress 
per  bq.  In. 
to 
Total  Strain. 

560 

0.0485| 

Percep- 
tible 

|     2668 

0.000082 

- 

32536600 

j 

Percep- 

] 

1120 

O.I095J 

tible 
after  one 

[   5335 

0.000185 

- 

28837800 

| 

hour 

J 

1680 

0.1675 

0.0015 

8003 

0.000284 

0.0000025 

28169000 

2240 

0.2240 

O.OO2O 

10670 

0.000379 

0.0000034 

28153000 

2800 

0.2805 

0.0027 

13338 

0.000475 

O.OOOOO42 

28O8OOOO 

3360 

03370 

0.0030 

16005 

0.000571 

O.OOOOO47 

28029000 

3920 

0.3930 

0.0040 

18673 

O.OOO666 

O.OOOOO68 

28037500 

4480 

0.4520 

0.0075 

21340 

0.000766 

O.OOOOI27 

27859000 

5040 

o-S^S 

0.0195 

24008 

0.000874 

0.0000330 

27469100 

394 


APPLIED  MECHANICS. 


Weights 
applied,  in 
Ibs. 

Exten- 
sion, in 
inches. 

Sets,  in 
inches. 

Weights 
per  Square 
Inch  of 
Section,  in 
Ibs. 

Strains,  in 
Fractions 
of  the 
Length. 

Sets,  in  Frac- 
tions of 
the  Length. 

Ratio  of  Str^is 
per  Sq   in. 
to 
Total  Strain. 

5600 
6l6o 

0.5980 
0.7600 

0.0490 
0.1545 

26676 
29343 

O.OOIOI4 
0.001288 

0.0000830 
0.0002619 

2630/700 
22781800 

6720 

I.3IOO 

320II 

0.002228 

— 

- 

In  10  min. 

1.3780 

0.6670 

0.002356 

0.0011305 

13587000 

In  12  min. 

1.3900 

— 

— 

- 

— 

— 

7280 

2-5310 

1.8125 

34678 

0.004290 

0.0030720 

8083500 

7840 
Repeated 
8400 

5.4060 
5-8750 
6.2190 

5.OOOO 
5.0625 
5-3750 

37346 
40013 

0.009163 
0.009957 

0.010166 

0.0084746 
0.0085805 
0.0091083 

4075700 
3936000 

In  i  hour 

6.9370 

— 

0.011758 

— 

— 

In  2  hours 

7.0000 

- 

- 

0.011866 

- 

_ 

In  3  hours 

7.0470 

- 

- 

0.011775 

- 

_ 

In  4  hours 

7.0500 

— 

— 

0.011950 

— 

_ 

In  5  hours 

7.0620 

_ 

_ 

0.011966 

_ 

_ 

In  6  hours 

7.0640 

- 

- 

0.011975 

- 

_ 

In  7  hours 

7.0940 

— 

— 

0.012025 

— 

_ 

In  10  hours 

7.0940 

- 

- 

0.012025 

- 

_ 

Left  over 

] 

night  ; 

next  morn- 

1 7.0000 

~ 

~ 

~ 

~ 

- 

ing 

j 

8960 

10.5620 

9.7500 

42681 

0.017900 

0.0165250 

2384400 

In  5  min. 

11.5000 

— 

0.019492 

— 

— 

In  10  min. 

11.7190 

_ 

_ 

0.019858 

— 

_ 

In  i  hour 

10.875 

- 

- 

0.0184333 

_ 

In  46  hours 

11.9370 

1  1  .OOOO 

- 

0.020233 

0.0186417 

_ 

9520 

12.6870 

11.6880 

45348 

0.021500 

0.0198083 

'2IO920O 

In  i  hour 

12.8100 

_ 

0.021708 

— 

_ 

In  2  hours 

12.8150 

_ 

_ 

0.021716 

- 

_ 

In  1  9  hours 

12.8150 

11.8150 

- 

0.021716 

0.0200250 

_ 

10080 

14.6250 

— 

48016 

0.024792 

— 

_ 

In  10  min. 
In  i  hour 

14.8440 

134370 

: 

0.025158 

0.0227750 

1936700 

In  1  1  hours 

14.8910 

- 

- 

0.025242 

- 

Next  morn- 
ing 

j  14.8750 

- 

- 

- 

- 

- 

10640 

_ 

_ 

_ 

— 

— 

_ 

In  10  min. 

20.6250 

I9-375° 

50684 

0.034958 

0.0328417 

1449800 

Repeated 

20.7780 

— 

— 

0.035217 

— 

_ 

In  7  hours 

20.7810 

— 

— 

0.035225 

— 

_ 

In  12  hours 

20.7810 

— 

— 

0.035225 

— 

_ 

Next  morn- 

i 

ing 

j-  20.7  500 

~ 

1  1  200 

53351 

TENSILE,   ETC.,   STRENGTH  OF   WROUGHT-IRON.        395 

With  this  last  weight  the  rod  broke  at  one  of  the  weldings, 
where  there  was  a  slight  defect ;  perhaps  a  rather  smaller  weight 
would  have  broken  it. 

In  connection  with  this  table  of  results,  the  following  ob- 
servations will  be  made  :  — 

i°.  The  breaking-strength  per  square  inch  in  this  case  was 
53351  Ibs.  per  square  inch. 

2°.  Permanent  set  began  with  very  small  loads. 

3°.  The  limit  of  elasticity  was  about  21000  Ibs.  per  square 
inch,  this  being  about  the  load  where  permanent  set  began  to 
increase  rapidly. 

4°.  The  modulus  of  elasticity  of  the  rod  in  this  case  was 
about  28000000  Ibs.  per  square  inch,  this  being  the  ratio  of  the 
stress  to  the  strain  under  loads  less  than  the  limit  of  elasticity. 

5°.  This  experiment  shows,  that,  under  moderate  loads,  the 
ratio  of  the  stress  to  the  strain  remains  much  more  nearly 
constant  with  wrought  than  with  cast  iron. 

6°.  The  other  experiment,  with  a  rod  50  feet  long,  gave  as 
modulus  of  elasticity  27691200  Ibs.  per  square  inch. 

The  remaining  experiments  of  Hodgkinson  on  wrought- 
iron  may  be  found  in  the  report  of  the  commissioners  already 
referred  to. 

Barlow 's  experiments  on  wrought-iron  will  not  be  detailed 
here,  except  only  to  say  that  he  tried  seven  experiments  to 
determine  the  modulus  of  elasticity  of  wrought-iron,  and  that 
he  obtained,  for  mean  extension  per  English  ton  per  square  inch, 
maximum,  0.0001082  ;  minimum,  0.0000841  ;  mean,  0.0000956 
of  the  length.  These  correspond  to  moduli  of  elasticity  re- 
spectively equal  to  20702400,  26634900,  and  23430900  Ibs.  per 
square  inch. 

The  results  obtained  by  experimenters  before  Hodgkin- 
son's  time  are  very  discordant  and  very  uncertain,  many  of 
them  attributing  to  wrought-iron  a  strength  far  greater  than 


396  APPLIED   MECHANICS. 

can  be  attained  at  the  present  time.  There  is  no  satisfactory 
evidence,  however,  to  show  that  the  wrought-iron  of  that  time 
was  any  better  than  (if  as  good  as)  that  made  at  the  present 
time ;  and  it  is  more  probable  either  that  there  were  errors  in 
making  the  tests,  or  else  that  the  supposed  iron  was  really  steel, 
and  possibly  brittle.  The  paucity  of  the  records,  and  the  im- 
possibility of  obtaining  the  details  of  the  tests,  render  any 
search  for  the  reason  in  any  special  cases  futile,  and  throw 
doubt  upon  the  greater  part  of  these  tests. 

Among  the  later  English  experimenters,  we  have  Sir  William 
Fairbairn.  An  account  of  his  tests  on  tensile  strength  will  be 
found  in  the  book  already  referred  to. 

One  of  the  most  prominent  English  experimenters,  and  one 
who  did  at  one  time  a  great  deal  towards  rendering  the  knowl- 
edge of  this  subject  more  accurate,  is  David  Kirkaldy.  His 
results  up  to  1866  are  detailed  in  his  book  entitled  "Experi- 
ments on  Wrought-iron  and  Steel,"  published  in  that  year. 

In  the  early  part  of  his  book  will  be  found  a  summary  of 
what  had  been  done  in  this  line  by  the  earlier  experimenters. 

Kirkaldy  tested  a  large  number  of  English  irons;  and  a 
summary  of  his  results  will  be  given  here,  together  with  his 
sixty-six  concluding  observations. 


TENSILE,   ETC.,   STRENGTH  OF   WROUGHT-IRON.          3Q7 


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TENSILE,   ETC.,   STRENGTH  OF   WROUGHT-IRON.        399 


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APPLIED   MECHANICS. 


SUMMARY   OF   KIRKALDY'S   RESULTS.  — IRON   PLATES. 


Names  of  Makers  or  Works. 

Lengthwise  or 
Crosswise. 

Breaking-  Weight  per 
Square  Inch  of  Area. 

Contrac- 
tion of 
Area  at 
Fracture, 
per  cent. 

Extreme 
Elonga- 
tion, 
per  cent. 

Original, 
Ibs. 

Fractured, 
Ibs. 

L. 
C. 
L. 
C. 
L. 
C. 
L. 
C. 
L. 
C. 
L. 
C. 
L. 
C. 
L. 
C. 
L. 
C. 
L. 
C. 
L. 
C. 
L. 
C. 
L. 
C. 
L. 
C. 
L. 
C. 
I!. 
C. 
L. 
C. 

58437 
55033 
58487 
54098 
52OOO 

50515 
56005 
46221 

52235 
46441 

54644 

49399 
55831 
50550 
56996 

5I25i 
55708 

49425 
52572 
50627 

5i245 
46712 

53559 
45677 
49120 

46755 
54847 
45585 
52362 

43036 
45997 
493  i  1 
53849 
48848 

83112 

68961 
70538 
59698 
64746 

57383 
68763 

53293 
61716 
50009 
66728 
54020 
67406 
55206 
66858 
56070 
65652 
54002 
62131 
55746 

59  '83 

52050 
62346 
48358. 
55472* 
50000 

62747 
47712 
61581 

4530° 
51140 
54842 
60522 

52252 

29.6 

20.1 
2O.O 
II.4 

19-7 
I2.I 
I7.8 
I3.2 

!5-3 

6.9 

18.1 
8.5 
17.2 
9.0 
15.0 
8.6 
14.9 
8.1 
15-4 
9-3 
13-1 

10.2 
144 
5.6 

"•3 

6-3 
12.5 
4.6 
15.0 

5-3 
10.4 

10.5 

II.O 

6.6 

17.0 

1  1-3 
10-9 

5-9 

13.2 

9-3 
14.1 
7.6 
u.6 

5-9 
11.6 
6.5 
12.5 

5-5 
13.0 

5-9 
10.7 

5-1 
8.6 
5.8 
8.9 
6.4 

"•5 

4.0 
8.0 
5-2 

II.  2 

4.6 
9.6 

2.8 

6.7 

7.0 

9-3 
4.6 

<O  Govan  <£>     

Bradley  &  Co.       S.  C  
Bradley  &  Co    L  F  

Bradley  &  Co      

Malinslee       Best    

Consett  Best  Best         

Consett  Best  Best    

Consett  Best  Best    

Thorneycroft  Best  Best    .... 
Snedshill  ^  Best  

Wells  Best  "^  Best 

Glasgow  Best  Boiler    ...         . 

KIRKALDY'S  SIXTY-SIX   CONCLUSIONS. 


401 


SUMMARY   OF   KIRKALDY'S    RESULTS.  —  IRON    PLATES.  —  Concluded. 


Names  of  Makers  or  Works. 

Lengthwise  or 
Crosswise. 

Breaking-  Weight  per 
Square  Inch  of  Area. 

Contrac- 
tion of 
Area  at 
Fracture, 
per  cent. 

Extreme 
Elonga- 
tion, 
per  cent. 

Original, 
Ibs. 

Fractured, 
Ibs. 

Glasgow  Best  Best 

L. 

53399 

59557 

10.6 

9.0 

C. 

41791 

436l4 

3-7 

2.6 

Wells  Best  ^  Best     

L. 

A.7A.IQ 

CI  CT 

67 

A    O 

C. 

TV  *f*W 

46630 

J    J    *• 

48348 

w./ 

4-9 

£\.*\J 

3-4 

Maker's  stamp  uncertain  .... 

L. 

47598 

53182 

9.8 

5-9 

C. 

40682 

43426 

5-o 

2-5 

K.  B.  M.     .  ...    A    ..  ,    ,    .    ,  ;  v 

L. 

46404 

51896 

IO.2 

6.1 

C. 

44764 

47891 

6.4 

4-3 

Lloyds,  Foster,  &  Co.,  Best  .     .     . 

L. 

44967 

49162 

8.7 

5-3 

C. 

44732 

48344 

6.9 

4.6 

Glasgow  Best  Best       . 

L. 

45626 

48208 

5-3 

4-3 

C. 

41340 

42430 

2.2 

2.4 

L. 

d.7777 

49816 

48 

T.6 

C. 

4/  /  /  j 

44355 

45343 

4>o 

3-4 

o*y 

2.1 

Mossend  Best  Best       .     .     < 

L. 

43433 

46038 

5-7 

3-3 

C. 

4H56 

43622 

4.9 

2.9 

Govan  Best     

L. 

43942 

45886 

6.0 

3-4 

C. 

39544 

40624 

2.6 

1.4 

Glasgow  Best  Scrap     .     .     .     .     , 

L. 

50844 

58412 

13.0 

10.5 

Fracture    1-12.  —  Light  gray  of  various  shades,  close  and  very  fine. 
Fracture  13-30.  —  Dullish  gray  of  various  shades,  close  and  generally  fine. 
Fracture  31-44.  —  Dull  gray,  generally  rather  coarse  and  irregular. 
Fracture  45-52.  —  Irregular,  generally  coarse  and  open. 
Fracture        53.  —  Close  and  fine. 


KIRKALDY'S  SIXTY-SIX   CONCLUSIONS. 

i°.  The  breaking-strain  does  not  indicate  the  quality,  as  hitherto 
assumed. 

2°.  A  high  breaking-strain  may  be  due  to  the  iron  being  of  superior 
quality,  dense,  fine,  and  moderately  soft,  or  simply  to  its  being  very 
hard  and  unyielding. 


402  APPLIED   MECHANICS. 

3°.  A  low  breaking-strain  may  be  due  to  looseness  and  coarseness 
in  the  texture,  or  to  extreme  softness,  although  very  close  and  fine  in 
quality. 

4°.  The  contraction  of  area  at  fracture,  previously  overlooked, 
forms  an  essential  element  in  estimating  the  quality  of  specimens. 

5°.  The  respective  merits  of  various  specimens  can  be  correctly 
ascertained  by  comparing  the  breaking-strain  jointly  with  the  contrac- 
tion of  area. 

6°.  Inferior  qualities  show  a  much  greater  variation  in  the  break- 
ing-strain than  superior. 

7°.  Greater  differences  exist  between  small  and  large  bars  in  coarse 
than  in  fine  varieties. 

8°.  The  prevailing  opinion  of  a  rough  bar  being  stronger  than  a 
turned  one,  is  erroneous. 

9°.  Rolled  bars  are  slightly  hardened  by  being  forged  down. 

10°.  The  breaking-strain,  and  contraction  of  area,  of  iron  plates  are 
greater  in  the  direction  in  which  they  are  rolled  than  in  a  transverse 
direction. 

11°.  A  very  slight  difference  exists  between  specimens  from  the 
centre  and  specimens  from  the  outside  of  crank-shafts. 

1 2°.  The  breaking-strain,  and  contraction  of  area,  are  greater  in  those 
specimens  cut  lengthways  out  of  crank-shafts  than  in  those  cut  cross-ways. 

13°.  The  breaking-strain  of  steel,  when  taken  alone,  gives  no  clew 
to  the  real  qualities  of  various  kinds  of  that  material. 

14°.  The  contraction  of  area  at  fracture  of  specimens  of  steel  must 
be  ascertained,  as  well  as  in  those  of  iron. 

15°.  The  breaking-strain,  jointly  with  the  contraction  of  area,  affords 
the  means  of  comparing  the  peculiarities  in  various  lots  of  specimens. 

1 6°.  Some  descriptions  of  steel  are  found  to  be  very  hard,  and  con- 
sequently suitable  for  some  purposes ;  whilst  others  are  extremely  soft, 
and  equally  suitable  for  other  uses. 

1 7°.  The  breaking-strain,  and  contraction  of  area,  of  puddled  steel 
plates,  as  in  iron  plates,  are  greater  in  the  direction  in  which  they  are 
rolled  ;  whereas  in  cast-steel  they  are  less. 

1 8°.  Iron,  when  fractured  suddenly,  presents  invariably  a  crystalline 
appearance ;  when  fractured  slowly,  its  appearance  is  invariably  fibrous. 


KIRKALDY'S  SIXTY- SIX   CONCLUSIONS.  403 

19°.  The  appearance  may  be  changed  from  fibrous  to  crystalline  by 
merely  altering  the  shape  of  the  specimen  so  as  to  render  it  more  liable 
to  snap. 

20°.  The  appearance  may  be  changed  by  varying  the  treatment  so 
as  to  render  the  iron  harder  and  more  liable  to  snap. 

21°.  The  appearance  may  be  changed  by  applying  the  strain  so 
suddenly  as  to  render  the  specimen  more  liable  to  snap  from  having  less 
time  to  stretch. 

22°.  Iron  is  less  liable  to  snap  the  more  it  is  worked  and  rolled. 

23°.  The  "skin,"  or  outer  part  of  the  iron,  is  somewhat  harder  than 
the  inner  part,  as  shown  by  appearance  of  fracture  in  rough  and  turned 
bars. 

24°.  The  mixed  character  of  the  scrap-iron  used  in  large  forgings 
is  proved  by  the  singularly  varied  appearance  of  the  fractures  of  speci- 
mens cut  out  of  crank-shafts. 

25°.  The  texture  of  various  kinds  of  wrought-iron  is  beautifully 
developed  by  immersion  in  dilute  hydrochloric  acid,  which,  acting  on 
the  surrounding  impurities,  exposes  the  metallic  portion  alone  for  exam- 
ination. 

26°.  In  the  fibrous  fractures  the  threads  are  drawn  out,  and  are 
Viewed  externally;  whilst  in  the  crystalline  fractures  the  threads  are 
snapped  across  in  clusters,  and  are  viewed  internally  or  sectionally.  In 
the  latter  cases,  the  fracture  of  the  specimen  is  always  at  right  angles 
to  the  length ;  in  the  former,  it  is  more  or  less  irregular. 

2  7°.  Steel  invariably  presents,  when  fractured  slowly,  a  silky,  fibrous 
appearance ;  when  fractured  suddenly,  the  appearance  is  invariably 
granular,  in  which  case  also  the  fracture  is  always  at  right  angles  to  the 
length ;  when  the  fracture  is  fibrous,  the  angle  diverges  always  more  or 
Jess  from  90°. 

28°.  The  granular  appearance  presented  by  steel  suddenly  fractured 
is  nearly  free  from  lustre,  and  unlike  the  brilliant  crystalline  appearance 
of  iron  suddenly  fractured  :  the  two  combined  in  the  same  specimen 
are  shown  in  iron  bolts  partly  converted  into  steel. 

29°.  Steel  which  previously  broke  with  a  silky,  fibrous  appearance  is 
changed  into  granular  by  being  hardened. 

30°.  The  little  additional  time  required  in  testing  those  specimens 


404  APPLIED   MECHANICS. 

whose  rate  of  elongation  was  noted,  had  no  injurious  effect  in  lessening 
the  amount  of  breaking-strain,  as  imagined  by  some. 

31°.  The  rate  of  elongation  varies  not  only  extremely  in  differeni 
qualities,  but  also  to  a  considerable  extent  in  specimens  of  the  same 
brand. 

32°.  The  specimens  were  generally  found  to  stretch  equally  through- 
out their  length  until  close  upon  rupture,  when  they  more  or  less  sud- 
denly draw  out ;  usually  at  one  part  only,  sometimes  at  two,  and  in  a  few 
exceptional  cases  at  three,  different  places. 

33°.  The  ratio  of  ultimate  elongation  may  be  greater  in  short  than 
in  long  bars  in  some  descriptions  of  iron,  whilst  in  others  the  ratio  is  not 
affected  by  difference  in  the  length. 

34°.  The  lateral  dimensions  of  specimens  form  an  important  ele- 
ment in  comparing  either  the  rate  of  or  the  ultimate  elongations,  —  a 
circumstance  which  has  been  hitherto  overlooked. 

35°.  Steel  is  reduced  in  strength  by  being  hardened  in  water,  while 
the  strength  is  vastly  increased  by  being  hardened  in  oil. 

36°.  The  higher  steel  is  heated  (without,  of  course,  running  the  risk 
of  being  burned),  the  greater  is  the  increase  of  strength  by  being 
plunged  into  oil. 

37°.  In  a  highly  converted  or  hard  steel  the  increase  in  strength 
and  in  hardness  is  greater  than  in  a  less  converted  or  soft  steel. 

38°.  Heated  steel,  by  being  plunged  into  oil  instead  of  water,  is  not 
only  considerably  hardened,  but  toughened,  by  the  treatment. 

39°.  Steel  plates  hardened  in  oil  and  joined  together  with  rivets  are 
fully  equal  in  strength  to  an  unjointed  soft  plate,  or,  the  loss  of  strength 
by  riveting  is  more  than  counterbalanced  by  the  increase  in  strength  by 
hardening  in  oil. 

40°.  Steel  rivets  fully  larger  in  diameter  than  those  used  in  riveting 
iron  plates  of  the  same  thickness,  being  found  to  be  greatly  too  small  for 
riveting  steel  plates,  the  probability  is  suggested  that  the  proper  propor- 
tion for  iron  rivets  is  not,  as  generally  assumed,  a  diameter  equal  to  the 
thickness  of  the  two  plates  to  be  joined. 

41°.  The  shearing-strain  of  steel  rivets  is  found  to  be  about  a  fourth 
less  than  the  tensile  strain. 

42°.  Iron  bolts  case-hardened  bore  a  less  breaking-strain  than  when 


KIRKALDY'S  SIXTY-SIX   CONCLUSIONS.  405 

wholly  iron,  owing  to  the  superior  tenacity  of  the  small  proportion  of 
steel  being  more  than  counterbalanced  by  the  greater  ductility  of  the 
remaining  portion  of  iron. 

43°.  Iron  highly  heated,  and  suddenly  cooled  in  water,  is  hardened, 
and  the  breaking-strain,  when  gradually  applied,  increased ;  but  at  the 
same  time  it  is  rendered  more  liable  to  snap. 

44°.  Iron,  like  steel,  is  softened,  and  the  breaking-strain  reduced, 
by  being  heated,  and  allowed  to  cool  slowly. 

45°.  Iron  subjected  to  the  cold-rolling  process  has  its  breaking- 
strain  greatly  increased  by  being  made  extremely  hard,  and  not  by  being 
*'  consolidated,"  as  previously  supposed. 

46°.  Specimens  cut  out  of  crankshaft  are  improved  by  additional 
hammering. 

47°.  The  galvanizing  or  tinning  of  iron  plates  produces  no  sensible 
effects  on  plates  of  the  thickness  experimented  on.  The  results,  how- 
ever, may  be  different  should  the  plates  be  extremely  thin. 

48°.  The  breaking-strain  is  materially  affected  by  the  shape  of  the 
specimen.  Thus,  the  amount  borne  was  much  less  when  the  diameter 
was  uniform  for  some  inches  of  the  length  than  when  confined  to  a 
small  portion,  —  a  peculiarity  previously  unascertained  and  not  even 
suspected. 

49°.  It  is  necessary  to  know  correctly  the  exact  conditions  uncler 
which  any  tests  are  made,  before  we  can  equitably  compare  results 
obtained  from  different  quarters. 

50°.  The  startling  discrepancy  between  experiments  made  at  the 
Royal  Arsenal  and  by  the  writer  is  due  to  the  difference  in  the  shape 
of  the  respective  specimens,  and  not  to  the  difference  in  the  two  testing- 
machines. 

51°.  In  screwed  bolts,  the  breaking-strain  is  found  to  be  greater 
when  old  dies  are  used  in  their  formation  than  when  the  dies  are  new, 
owing  to  the  iron  becoming  harder  by  the  greater  pressure  required  in 
forming  the  screw-thread  when  the  dies  are  old  and  blunt  than  when 
new  and  sharp. 

52°.  The  strength  of  screw-bolts  is  found  to  be  in  proportion  to 
their  relative  areas ;  there  being  only  a  slight  difference  in  favor  of  the 
smaller  compared  with  the  larger  sizes,  instead  of  the  very  material 
difference  previously  imagined. 


406  APPLIED   MECHANICS. 

53°.  Screwed  bolts  are  not  necessarily  injured,  although  strained 
nearly  to  their  breaking-point. 

54°.  A  great  variation  exists  in  the  strength  of  iron  bars  which  have 
been  cut  and  welded  :  whilst  some  bear  almost  as  much  as  the  uncut 
bar,  the  strength  of  others  is  reduced  fully  a  third. 

55°.  The  welding  of  steel  bars,  owing  to  their  being  so  easily  burned 
by  slightly  over-heating,  is  a  difficult  and  uncertain  operation. 

56°.  Iron  is  injured  by  being  brought  to  a  white  or  welding  heat 
if  not  at  the  same  time  hammered  or  rolled. 

57°.  The  breaking-strain  is  considerably  less  when  the  strain  is 
applied  suddenly,  instead  of  gradually,  though  some  have  imagined  that 
the  reverse  is  the  case. 

58°.  The  contraction  of  area  is  also  less  when  the  strain  is  suddenly 
applied. 

59°.  The  breaking-strain  is  reduced  when  the  iron  is  frozen.  With 
the  strain  gradually  applied,  the  difference  between  a  frozen  and  un- 
frozen bolt  is  lessened  as  the  iron  is  warmed  by  the  drawing-out  of  the 
specimen. 

60°.  The  amount  of  heat  developed  is  considerable  when  the 
specimen  is  suddenly  stretched,  as  shown  in  the  formation  of  vapor 
from  the  melting  of  the  layer  of  ice  on  one  of  the  specimens,  and  also 
by  the  surface  of  others  assuming  tints  of  various  shades  of  blue  and 
orange,  not  only  in  steel,  but  also,  although  in  a  less  marked  degree,  in 
iron. 

61°.  The  specific  gravity  is  found  generally  to  indicate  pretty  cor- 
rectly the  quality  of  specimens. 

62°.  The  density  of  iron  is  decreased  by  the  process  of  wire-drawing, 
and  by  the  similar  process  of  cold-rolling,  instead  of  increased,  as  pre- 
viously imagined. 

63°.  The  density  in  some  descriptions  of  iron  is  also  decreased  by 
additional  hot-rolling  in  the  ordinary  way :  in  others  the  density  is  very 
slightly  increased. 

64°.  The  density  of  iron  is  decreased  by  being  drawn  out  under  a 
tensile  strain,  instead  of  increased,  as  believed  by  some. 

65°.  The  most  highly  converted  steel  does  not,  as  some  may  sup- 
pose, possess  the  greatest  density. 


TESTS   OF  COMMANDER   BEARDSLEE,  407 

66°.  In  cast-steel  the  density  is  much  greater  than  in  puddled  steel, 
which  is  even  less  than  in  some  of  the  superior  descriptions  of  wrought- 
iron. 


TESTS   OF   COMMANDER   BEARDSLEE. 

One  of  the  most  valuable  sets  of  tests  of  wrought-iron 
is  that  obtained  by  committees  D,  H,  and  M  of  the  Board 
appointed  by  the  United-States  Government  to  test  iron  and 
si  eel ;  the  special  duties  of  these  committees  being  to  test  such 
iron  as  would  be  used  in  chain-cable,  and  the  chain-cable  itself. 
The  chairman  of  these  three  committees,  which  were  consoli- 
dated into  one,  was  Commander  L.  A.  Beardslee  of  the  United- 
States  Navy.  The  full  account  of  the  tests  is  to  be  found  in 
Executive  Document  98,  45th  Congress,  second  session  ;  and 
an  abridged  account  of  them  was  published  by  William  Kent, 
as  has  been  already  mentioned. 

The  samples  of  bar-iron  tested  were  round,  and  varied  from 
one  inch  to  four  inches  in  diameter.  The  mills  from  which  the 
samples  tested  came  were  as  follows  :  — 

Bentoni Pennsylvania. 

Burden  and  Sons New  York. 

Burgess Ohio. 

Catasauqua Pennsylvania. 

New- Jersey  Iron  and  Steel  Company     .     .     .  New  Jersey. 

Niles  Iron  Company Ohio. 

Pembroke Massachusetts. 

Pencoyd Pennsylvania. 

Phoenix Pennsylvania. 

Sligo  .     ,;  .   •;    . '  r."  ;;;  . -;:  ;  •  't-f\j->:  ."'V    .  Pennsylvania. 

Tamaqua Pennsylvania. 

Tredegar       •     •     • Virginia. 

Trego  and  Thompson Maryland. 

Wyeth  Brothers Maryland. 


408  APPLIED   MECHANICS. 

Certain  conclusions  which  they  reached  refer  to  all  kinds 
of  wrought-iron,  and  will  be  given  here  before  giving  a  table  of 
the  results  of  the  tests. 

i°.  Kirkaldy  considers  the  breaking-strength  per  square 
inch  of  fractured  area  as  the  main  criterion  by  which  to  deter- 
mine the  merits  of  a  piece  of  iron  or  steel.  Commander 
Beardslee,  on  the  other  hand,  thinks  that  a  better  criterion  is 
what  he  calls  the  "tensile  limit;"  i.e.,  the  maximum  load  the 
piece  sustains  divided  by  the  area  of  the  smallest  section  when 
that  load  is  on,  i.e.,  just  before  the  load  ceases  to  increase  in 
the  testing-machine. 

2°.  Kirkaldy  had  already  called  attention  to  the  fact  that 
the  tensile  strength  of  a  specimen  is  very  much  affected  by  its 
shape,  and  that,  in  a  specimen  where  the  shape  is  such  that 
the  length  of  that  part  which  has  the  smallest  cross-section  is 
practically  zero  (as  is  the  case  when  a  groove  is  cut  around 
the  specimen),  the  breaking-strength  is  greater  than  it  is  when 
this  portion  is  long ;  the  excess  being  in  some  cases  as  much 
as  33  per  cent. 

Commander  Beardslee  undertook,  by  actually  testing  speci- 
mens whose  smallest  areas  varied  in  length,  to  determine  what 
must  be  the  least  length  of  that  part  of  the  specimen  whose 
cross-section  area  is  smallest,  in  order  that  the  tensile  strength 
may  not  be  greater  than  with  a  long  specimen.  The  conclusion 
reached  was,  that  no  test-piece  should  be  less  than  one-half  inch 
in  diameter,  and  that  the  length  should  never  be  less  than  four 
diameters  ;  while  a  length  of  five  or  six  diameters  is  necessary 
with  soft  and  ductile  metal  in  order  to  insure  correct  results. 
The  following  results  of  testing  steel  are  given  in  Mr.  Kent's 
book,  as  confirming  the  same  rule  in  the  case  of  steel.  The 
tests  were  made  upon  Bessemer  steel  by  Col.  Wilmot  at  the 
Woolwich  arsenal. 


TESTS   OF  COMMANDER  BEARDSLEE. 


409 


Tensile  Strength. 

Pounds  per 
Square  Inch. 

Highest  .     .     .  ".  ^l' 
Lowest 

162974 

Average.     .     .     .     * 
Highest.     .  -i     .  v*:: 
T  owpst 

I3°49° 
153677 
123165 

Average  .     .     .     .  .a 

1U3255 
114460 

3°.  Commander  Beardslee  also  noticed  that  rods  of  certain 
diameters  of  the  same  kind  of  iron  bore  less  in  proportion  than 
rods  of  other  diameters  ;  and,  after  searching  carefully  for  the 
reason,  he  found  it  to  lie  in  the  proportion  between  the  diam- 
eter of  the  rod  and  the  size  of  the  pile  from  which  it  is 
rolled.  The  following  examples  are  given  :  — 

> 

i-J-in.  diameter,  6.62%  of  pile,  56543  Ibs.  per  sq.  in.  tensile  strength. 


I* 
If 
It 


8.i8% 

56478  ' 

9.90% 

"    54277  " 

11.78% 

"    53550  ;" 

7.68% 

"    56344  " 

8.90% 

"    55018  " 

10.22% 

"    54034  " 

II-63% 

"    51848  " 

He  therefore  claims,  that,  in  any  set  of  tests  of  round  iron, 
it  is  necessary  to  give  the  diameter  of  the  rod  tested,  and  not 
merely  the  breaking-strength  per  square  inch. 

4°.  He  gives  evidence  to  show,  that  if  a  bar  is  under-heated, 
it  will  have  an  unduly  high  tenacity  and  elastic  limit ;  and  that 
if  it  is  over-heated,  the  reverse  will  be  the  case. 


4io 


APPLIED   MECHANICS. 


5°.  The  discovery  was  made  independently  by  Commander 
Beardslee  and  Professor  Thurston,  that  wrought-iron,  after 
having  been  subjected  to  its  ultimate  tensile  strength  without 
breaking  it,  would,  if  relieved  of  its  load  and  allowed  to  rest, 
have  its  breaking-strength  and  its  limit  of  elasticity  increased. 

He  tried  a  considerable  number  of  experiments,  studying 
the  action  of  this  law  under  different  periods  of  rest,  from  I 
minute  to  3  days  and  upwards ;  and  a  great  deal  of  valuable 
information  is  given  in  the  tables  of  the  report. 

The  most  characteristic  table  is  the  following  :  — 


EFFECT  OF  EIGHTEEN   HOURS'  REST   ON   IRONS   OF  WIDELY   DIFFER. 
ENT   CHARACTERS. 


Ultimate  Strength 

per  Square  Inch. 

r>          i 

First 

Second 

xvernarics. 

Strain. 

Strain. 

Boiler  iron     .     .     . 

48600 

56500 

Not  broken. 

it        it 

49800 

57000 

Broken  -\ 

it        it 

49800 

58000 

Broken  I  Average  gain, 

"        "... 

48100 

54400 

Broken   |                    15.8%. 

it        it 

48150 

5555° 

Broken  J 

Contract  chain  iron, 

50200 

54000 

Broken         "j 

ti           it        n 

50250 

53200 

Not  broken   1   Average 

it           it        tt 

SO?00 

55300 

Not  broken   ^          gain, 

if           ti        tt 

49600 

52900 

Not  broken   I              6.4%. 

tt           it        ti 

51200 

52800 

Not  broken  J 

Iron  K     .     .     .    ;<  , 

it     n 

58800 
59000 

64500 
65800 

Broken  ^ 
-r,    ,           Average  sain, 
Broken   j*                         ? 

it     n 

56400 

60600 

Broken  J                      94%' 

TESTS  OF  COMMANDER  BEARDSLEE. 


His  experiments  show  that  the  increase  is  in  irons  of  a 
fibrous  and  ductile  nature,  rather  than  in  brittle  and  steely  ones  : 
hence  the  latter  class  would  be  but  little  benefited  by  the  action 
of  this  law. 

The  following  table  of  results  is  given  in  the  report ;  showing 
how  the  strength  per  square  inch  varies  with  the  diameter  of 
the  piece,  in  consequence  of  the  amount  of  reduction  in  the 
rolls. 


Diameter,  in  inches. 

Name  of  Iron. 

Strength 
per 
Square 
Inch  of 
Original 
Area. 

Elastic 
Limit 
per 
Square 
Inch  of 
Original 
Area. 

Diameter,  in  inches.  1 

Name  of  Iron. 

Strength 
per 
Square 
Inch  of 
Original 
Area. 

Elastic 
Limit 
per 
Square 
Inch  of 
Original 
Area. 

Diameter,  in  inches. 

Name  of  Iron. 

Strength 
per 
Square 
Inch  of 
Original 
Area. 

Elastic 
Limit 
per 
Square 
Inch  of 
Original 
Area. 

"i 

F 

59885 

I 

D 

52900 

Ii 

D 

57977 

31996 

1 

F 

54090 

40980 

I 

F 

52819 

32267 

Ii 

P 

55782 

35596 

i 

C 

62700 

I 

F 

51400 

34600 

Ii 

Px 

56334 

33921' 

\ 

C 

59000 

ii 

K 

60458 

37344 

Ii 

N 

56478 

33251 

i 

C 

57700 

i* 

D 

59582 

33597 

Ii 

Fxl 

55253 

34784 

i 

C 

55400 

ii 

C 

57470 

31900 

Ii 

D 

55550 

28166 

\ 

F 

52275 

39126 

i* 

Fxl 

56434 

34682 

Ii 

E 

53893 

32712 

I 

F 

55450 

ii 

P 

57498 

413" 

Ii 

Fx2 

55132 

38603 

f 

F 

52050 

4 

N 

56143 

32267 

Ii 

Fx3 

53247 

32520 

I 

F 

57660 

i* 

Fx2 

55927 

37250 

Ii 

A 

53897 

27643 

\ 

F 

51546 

35933 

ii 

E 

53097 

33549 

Ii 

M 

53752 

- 

\ 

F 

50630 

33931 

ii 

Fx3 

54644 

34695 

Ii 

M 

54090 

- 

K 

61727 

^ 

D 

54687 

28166 

Ii 

F 

52970 

32075 

D 

61115 

33486 

ii 

A 

53900 

26787 

Ii 

F 

52729 

39608 

0 

57363 

37415 

ii 

F 

53850 

33457 

Ii 

M 

53022 

- 

Fxl 

55768 

34729 

ii 

0 

53035 

32410 

Ii 

F 

52620 

33220 

P 

57807 

39230 

ii 

F 

50149 

35493 

Ii 

0 

50040 

30730 

A 

54690 

34881 

ii 

F 

52267 

32019 

iA 

P 

545i8 

35898 

Fx2 

56790 

36885 

^ 

K 

59461 

36501 

•if 

M 

58926 

37548 

Fx3 

53915 

36336 

4 

P 

56876 

36868 

ii 

M 

57649 

385/8 

F 

5I92I 

31300 

ii 

C 

57897 

32469 

if 

D 

58021 

32152 

412 


APPLIED   MECHANICS. 


Diameter,  in  inches. 

Name  of  Iron. 

Strength 
per 
Square 
Inch  of 
Original 
Area. 

Elastic 
Limit 
per 
Square 
Inch  of 
Original 
Area. 

I  Diameter,  in  inches. 

? 

1 
CS 

£ 

Strength 
per 
Square 
Inch  of 
Original 
Area. 

Elastic 
Limit 
per 
Square 
Inch  of 
Original 
Area. 

Diameter,  in  inches. 

Name  of  Iron. 

Strength 
per 
Square 
Inch  of 
Original 
Area. 

Elastic 
Limit 
per 
Square 
Inch  of 
Original 
Area. 

if 

K 

55790 

3I034 

ii 

M 

54095 

35544 

if 

Fx2 

53438 

35870 

if 

C 

54949 

3I03° 

ii 

E 

54544 

33027 

if 

H 

523H 

29364 

if 

M 

54373 

35820 

it 

P 

52868 

29636 

if 

E 

5!946 

27695 

if 

N 

54277 

33622 

I* 

M 

53512 

- 

if 

0 

52401 

34012 

it 

Fxl 

52968 

33275 

it 

M 

52941 

if 

F 

52163 

33907 

if 

Fx3 

52733 

346o6 

ii 

Fx3 

52819 

34840 

if 

Q 

51205 

33318 

if 

E 

52254 

2593° 

ii 

Fxl 

53491 

3"4307 

if 

F 

50529 

35390 

if 

A 

53557 

33650 

ii 

M 

52736 

349oi 

if 

F 

50970 

33625 

it 

P 

52556 

30802 

ii 

N 

53555 

34690 

if 

C 

49030 

31099 

If 

F 

52537 

34469 

ii 

C 

52700 

3588o 

itt 

K 

56595 

38310 

If 

F 

52339 

39103 

ii 

H 

52462 

29992 

iii 

B 

54181 

If 

M 

530i6 

35379 

4 

D 

52155 

27708 

iH 

J 

54H4 

If 

Fx2 

5H87 

359" 

ii 

A 

51884 

28794 

itt 

B 

52895 

33H5 

If 

F 

51296 

31992 

ii 

F 

51994 

32054 

itt 

E 

52120 

35549 

If 

O 

5°594 

34940 

ii 

0 

509^9 

32312 

itt 

G 

57789 

34160 

I-fc 

P 

53345 

_."  A- 

ii 

F 

5H56 

34591 

itt 

C 

49821 

33i84 

1^ 

E 

53944 

32542 

ii 

Fx2 

51481 

349  i  7 

if 

K 

57874 

r& 

G 

53238 

32534 

ii 

J 

5I047 

- 

if 

Px 

54212 

33908 

«ft 

B 

52287 

324II 

ii 

M 

49292 

32597 

if 

C 

544io 

31354 

»# 

C 

51756 

32655 

if 

N 

56344 

35889 

if 

P 

52844 

33842 

«* 

J 

50400 

- 

if 

K 

57132 

35026 

if 

Fxl 

53846 

36573 

i* 

M 

57052 

38417 

if 

M 

57402 

35701 

if 

H 

538oo 

27856 

I* 

K 

57317 

33412 

T&. 

1  8 

P 

55634 

33522 

if 

N 

55oi8 

34283 

i\ 

D 

56505 

32496 

If 

C 

•  56227 

33207 

if 

D 

53472 

31892 

i\ 

M 

55466 

34780 

If 

Px 

54689 

33427 

if 

J 

53264 

- 

ii 

M 

55i3i 

33771 

If 

A 

54334 

32163 

if 

D 

52699 

27817 

ri 

P 

54159 

33  HO 

If 

D 

53695 

30087 

if 

Fx3 

53154 

35323 

ii 

M 

54540 

- 

If 

Fx3 

53339 

33540 

if 

E 

51606 

26541 

tt 

C 

55404 

34770 

If 

Fxl 

53537 

34335 

if 

A 

51509 

29404 

it 

E 

55415 

32869 

If 

D 

536i4 

30664 

if 

F 

50690 

32229 

ii 

M 

54816 

347i6 

If 

J 

52748 

- 

if 

G 

50395 

36254 

i* 

Px 

54354 

34617 

if 

E 

52675 

33745 

if 

C 

50312 

30852 

TESTS   OF  COMMANDED  BEARDSLEE. 


413 


Diameter,  in  inches. 

Name  of  Iron. 

Strength 
per 
Square 
Inch  of 
Original 
Area. 

Elastic 
Limit 
per 
Square 
Inch  of 
Original 
Area. 

Diameter,  in  inches. 

Name  of  Iron. 

Strength 
per 
Square 
Inch  of 
Original 
Area. 

Elastic 
Limit 
per 
Square 
Inch  of 
Original 
Area. 

Diameter,  in  inches. 

? 

M 
1 

Strength 
per 
Square 
Inch  of 
Original 
Area. 

Elastic 
Limit 
per 
Square 
Inch  of 
Original 
Area. 

If 

F 

50547 

35954 

i| 

F 

49355 

32855 

2 

0 

48249 

3HI3 

If 

Fx2 

S23M 

35320 

i| 

F 

48670 

23250 

2 

D 

49146 

33068 

If 

E 

49816 

31214 

ig 

0 

47478 

30842 

2 

D 

46151 

36050 

It 

F 

49738 

28907 

lit 

M 

5M74 

- 

*h 

M 

5^559 

- 

If 

0 

50129 

32271 

lit 

M 

5*707 

- 

2lV 

M 

49422 

- 

lit 

K 

56577 

- 

lit 

M 

51242 

- 

2* 

M 

50481 

- 

III 

B 

53655 

- 

2 

K 

60213 

3J44i 

2i 

M 

51225 

- 

III 

C 

50969 

30814 

2 

K 

57567 

30839 

2^ 

A 

48382 

30459 

ili 

G 

503  10 

33565 

2 

Px 

529*4 

31198 

2-& 

M 

51666 

- 

i  ft 

E 

50307 

29767 

2 

M 

52820 

- 

2i 

M 

51530 

- 

ill 

J 

48953 

2 

M 

49164 

- 

2* 

M 

51296 

- 

tf 

K 

55803 

31031 

2 

E 

51818 

27318 

H 

F 

48812 

- 

if 

C 

54447 

32334 

2 

P 

51684 

33*04 

2* 

F 

48894 

- 

ii 

D 

53IO° 

32074 

2 

P 

50834 

31878 

H 

F 

49164 

31966 

if 

N 

54004 

33610 

2 

N 

52127 

32461 

2i 

F 

49290 

32163 

i| 

Fxl 

52875 

35641 

-7 

N 

51370 

32460 

4 

P 

46866 

28241 

Ti 

1  6 

Fx3 

5336i 

35032 

2 

Fxl 

52011 

34702 

H 

F 

47344 

29758 

If 

P 

52505 

32312 

2 

C 

5"53 

29335 

2* 

F 

48475 

28932 

It 

E 

50880 

27100 

2 

D 

51146 

28567 

H 

F 

47428 

29941 

If 

D 

5H59 

27816 

2 

P 

49872 

29953 

2f 

F 

46446 

26333 

if 

Px 

51762 

32261 

2 

Fx2 

50000 

36184 

3 

F 

47761 

26400 

i| 

M 

50363 

2 

Fx3 

50763 

33172 

3i 

F 

47014 

24591 

if 

A 

50584 

28713 

2 

A 

50171 

28983 

3* 

F 

47000 

24961 

I  8 

F 

5I039 

33067 

2 

F 

48596 

27634 

3i 

F 

46667 

23636 

I«C 

Fx2 

5"59 

33970 

2 

F 

47812 

35864 

4 

F 

46322 

23430 

If 

F 

49744 

35615 

2 

F 

47569 

28792 

414 


APPLIED   MECHANICS. 


TENSILE    TESTS    MADE   SUBSEQUENTLY   AT    THE   WATERTOWN 

ARSENAL. 

Here  will  next  be  given,  in  tabulated  form,  the  results  of  a 
number  of  tensile  tests  made  on  the  government  machine  at  the 
Watertown  Arsenal. 

The  following  tables  of  results  on  rolled  bars,  from  the  Elmira 
Rolling-Mill  Company  (mark  L)  and  from  the  Passaic  Rolling- 
Mills  (mark  S),  are  given  in  Executive  Document  12,  4Jth  Con- 
gress, 1st  session,  and  in  Executive  Document  I,  4Jth  Congress, 
2d  session. 

SINGLE   REFINED   BARS. 


« 

c 

0 

•a 

rt 
§ 

Sectional  Area,  in 
square  inches. 

.5  £ 

a 
.*"  §• 

S* 

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P  • 

£ 

c 
c«£ 

1j 

bC  -g 

*0 

1^ 

li 

s< 

Appearance  of 
Fracture. 

Modulus  of  Elas- 
ticity at  Load  of 
20000  Lbs.  per 
Square  Inch. 

J^ 

3^ 

£  s 

u  — 

L   i 

3.06 

28500 

52710 

l8.4 

33-3 

95 

5 

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L    2 

3.06 

29500 

53630 

16.4 

36.0 

92 

8 

27826036 

L   3 

3.06 

29000 

52090 

21.4 

34-6 

95 

5 

28419182 

L   4 

3.06 

29000 

5  '440 

I5.0 

20.3 

90 

10 

30888030 

L   5 

6.46 

27500 

5°5°° 

14-5 

27.6 

95 

5 

27826036 

L   6 

6.40 

27500 

50530 

17-3 

22.3 

70 

30 

27118644 

L   7 

6.39 

27000 

50200 

1  8.0 

22.5 

95 

5 

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- 

51667 

22.0 

36.0 

70 

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Round. 

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- 

50844 

I6.3 

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85 

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L   10 

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53062 

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28500 

48640 

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28000 

50390 

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3-05 

28500 

47050 

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26000 

49700 

17.1 

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6.40 

24000 

49280 

15-7 

17.7 

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26490066 

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6.41 

24500 

48740 

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80 

-  20 

28119507 

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3-17 

24600 

49680 

19-5 

32.0 

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27972027 

Round. 

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25870 

49338 

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29357798 

« 

Cinder 

S    20 

3-*7 

24920 

48864 

18.4 

37-o 

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at  centre 

27729636 

H 

DOUBLE   REFINED   BARS. 


415 


DOUBLE    REFINED    BARS. 


1 
.  § 
3 

Sectional  Area,  in 
square  inches. 

c  v 

jjj 

ill 

Ultimate  Strength, 
in  Ibs.,  per 
Square  Inch. 

<§ 

Jf 

Contraction  of 
Area,  %. 

Appearance  of 
Fracture. 

Modulus  of  Elas- 
ticity at  Load  of 
20000  Ibs.  per 
Square  Inch. 

J* 

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3.06 

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0 

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L  202 

3-03 

30000 

52650 

16.2 

20.  6 

85 

15 

34042553 

L  203 

3-o6 

32500 

535oo 

16.5 

27-5 

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28169014 

L  204 

3.06 

32500 

54480 

15-4 

24.8 

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0 

29090909 

L  205 

6-33 

27000 

51230 

17.8 

24.2 

80 

20 

28119507 

L  206 

6-34 

27500 

50500 

17.6 

21.  1 

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Slightly 

29629629 

L  207 

6-34 

27OOO 

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21.4 

31-9 

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L  208 

3.20 

- 

50156 

22.7 

43-o 

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shaped 

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L  209 

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L  2IO 

3.20 

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50188 

19.9 

43-o 

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28985507 

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S  211 

3-05 

29500 

51150 

22.0 

31.5 

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32989690 

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3-05 

28500 

51110 

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36.1 

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29500 

51860 

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26446280 

S  215 

6.31 

27500 

50980 

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23.6 

95 

5 

29357798 

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6.38 

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20.7 

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28070175 

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28070175 

The  moduli  of  elasticity  had  not  been  computed  in  the 
report,  but  have  been  computed  in  these  tables  from  the  elon- 
gations under  a  load  of  20000  Ibs.  per  square  inch  in  each  case, 
as  recorded  in  the  details  of  the  tests. 

In  these  reports  are  also  to  be  found  tensile  tests  of  iron 
from  other  companies,  as  the  Detroit  Bridge  Company,  the 
Phoenix  Company,  the  Pencoyd  Company,  etc.  Some  of  these 


416 


APPLIED   MECHANICS. 


tests  were  made  to  determine  the  effect  of  rest  upon  the  bar 
after  it  had  been  strained  to  its  ultimate  strength,  also  to 
determine  the  strength  after  annealing.  The  following  table 
shows  these  latter  results : 


. 

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^^U               3^X13X13                J2U33C 

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(  since  or 

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2 

s    %          & 

it      oo'oo1              ^oo                oo'           r^.   oo         oo" 

DOUBLE  REFINED  BARS. 


417 


Some  tests  were  made  to  determine  the  values  of  the 
modulus  of  elasticity  of  the  same  iron  for  tension  and  for 
compression ;  and  these  were  found  experimentally  to  be 
almost  identical,  as  was  to  be  expected.  For  these  tests  the 
student  is  referred  to  the  reports  themselves  ;  and  only  cer- 
tain tests  on  eye-bars  of  the  Phoenix  Company  will  be 
appended  here.  Some  of  them  were  tested  only  for  modulus, 
of  elasticity,  which  are  not  calculated  in  the  report,  but  are 
given  here. 


Arsenal 
Num- 
ber. 

Outside 
Length, 
inches. 

Gauged 
Length, 
inches. 

Sectional 
Area, 
square 
inches. 

Modulus 
of 
Elasticity. 

Ultimate 
Strength, 
Ibs.,  per 
Sq.  In. 

Con- 
traction 
of  Area 
at  Frac- 
ture, %. 

5°9 

95-9° 

70 

0.765 

25309000 

510 

95-90 

70 

0.765 

27035000 

511 

6  7-  75 

5° 

1.478 

22391000 

40600 

16.8 

512 

6  7-  75 

5° 

15.090 

25075000 

5T3 

67.80 

5° 

1.940 

24727000 

39480 

13.9 

5J4 

118.75 

90 

x-959 

24590000 

5J5 

118.75 

90 

J-959 

25993000 

5i6 

96.05 

75 

2-954 

25388000 

5T7 

109.72 

75 

3.800 

25969000 

518 

96.05 

75 

5-I03 

23338000 

46720 

8.1 

Quite  a  number  of  tests  of  the  iron  of  different  American 
companies  are  to  be  found  in  the  "  Report  on  the  Progress  of 
Work  on  the  Cincinnati  Southern  Railway,"  by  Thomas  D. 
Lovett,  Nov.  i,  1875. 

For  these  the  student  is  referred  to  the  report  named. 


APPLIED  MECHANICS. 


SPECIAL    MODULUS    OF    ELASTICITY. 

In  connection  with  Hodgkinson's  experiment  on  a  bar  50 
feet  long,  attention  was  called  to  the  fact  that  the  ratio  of  the 
stress  to  the  total  strain,  which  is  often  called  the  modulus  of 
elasticity,  remained  more  nearly  constant  in  wrought  than  in 
cast  iron  for  loads  below  the  limit  of  elasticity,  and  that  after 
passing  this  limit  it  decreases  rapidly,  as  the  strain  increases 
much  more  rapidly  than  the  stress. 

This  could  be  represented  graphically  by  drawing  a  curve, 
having  for  abscissae  of  its  several  points  the  loads  per  square 
inch  applied  to  the  piece,  and  for  ordinates  the  elongations  or 
shortenings.  Such  a  curve  would,  in  the  case  of  wrought-iron, 
be  nearly  a  straight  line  up  to  the  limit  of  elasticity,  and  then 
would  rise  very  rapidly. 

If,  now,  we  should  plot  another  curve  with  the  same  abscissas 
as  before,  but  having  for  ordinates  the  permanent  sets  under 
the  respective  loads,  this  curve  would  start  from  the  axis  of 
abscissae  at  the  limit  of  elasticity,  and  rise  rapidly  from  that 
point  on. 

Again  :  if  another  curve  be  drawn,  having  the  same  abscissae 
as  before,  but  having  for  ordinates  in  every  case  the  differences 
of  the  ordinates  of  the  other  two  curves,  such  a  curve  would 
represent,  by  its  ordinate  at  any  point,  the  recoil  or  the  stretch 
of  the  piece  over  and  above  the  permanent  set ;  and  this 
ordinate  would,  in  the  absence  of  further  experiment,  appear 
to  represent  the  amount  it  would  stretch  if  the  load  were 
removed  and  immediately  applied  again,  over  and  above  that 
part  of  the  stretch  which  did  not  disappear  on  the  removal 
of  the  load.  The  modulus  of  elasticity  computed  by  means  of 
the  elongation  above  described,  considered  as  the  elongation 
of  the  piece,  has  been  given  the  name  "  special  modulus  of 
elasticity"  by  Col.  Rosset  of  the  arsenal  at  Turin  ;  and  he  has 
called  attention  to  the  fact,  that  in  the  case  of  wrought-iron, 


WROUGHT-IRON  AND   STEEL   EYE-BARS  419 


and  also  of  steel,  this  special  modulus  of  elasticity  is  nearly 
constant,  even  though  the  load  applied  be  far  above  the  limit 
of  elasticity. 

This  can  be  graphically  shown  from  the  fact  that  the  third 
of  the  above-mentioned  curves,  if  plotted,  will  be  nearly  a 
straight  line  almost  up  to  the  point  of  fracture  in  the  case  of 
wrought-iron  and  steel. 

Moreover,  this  special  modulus  of  elasticity  is  the  real 
modulus  of  elasticity  of  the  material,  and  it  is  this  that  is  meant 
whenever  any  values  of  the  modulus  of  elasticity  are  given  in 
connection  with  the  tests  made  on  the  Government  machine  at 
Watertown  Arsenal. 


EXAMPLES. 

1.  Plot  the  curves  referred  to  above,  for  Hodgkinson's  test  of  a 
wrought-iron  rod  50  feet  long,  recorded  on  p.  391. 

2.  Do  the  same  for  the  table  of  experiments  on  cast-iron  bars  10 
feet  long,  recorded  on  p.  360. 


WROUGHT-IRON    AND    STEEL    EYE-BARS. 

In  the  report  of  the  Government  tests  for  1886  is  given 
the  following  table  of  tensile  tests  of  wTought-iron  and  steel 
eye-bars.  The  wrought-iron  ones  were  furnished  by  the  Gen- 
eral  Manager  of  the  Boston  and  Maine  Railroad,  and  the  steel 
ones  by  the  Chief  Engineer  for  the  American  Committee  of 
the  Statue  of  Liberty. 


420 


APPLIED   MECHANICS. 


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WROUGHT-IRON  PLATE. 


421 


WROUGHT-IRON    PLATE. 

Some  tests  of  wrought-iron  plate  have  already  been  quoted, 
viz.,  Kirkaldy's ;  and  the  following  table  is  added,  giving  some 
tests  of  wrought-iron  plate  and  bars  made  on  the  Govern- 
ment testing-machine  at  Watertown  in  1883  and  1884  for  the 
Supervising  Architect  at  Washington,  D.C. 


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422  APPLIED  MECHANICS. 

BRANDS    OF    PLATES. 

The  only  safe  way,  in  ordering  plates  for  a  boiler,  as  in  or- 
dering iron  for  any  construction,  is  to  prescribe  the  tests  they 
shall  stand,  as  the  same  brands  are  used  by  different  makers 
to  denote  very  different  qualities  of  iron.  Indeed,  the  fact 
that  steel  is  very  largely  taking  the  place  of  iron  in  the  making 
of  the  better  class  of  boilers  is  rendering  it  gradually  more  dif- 
ficult to  obtain  good  wrought-iron  boiler-plate. 

As  a  rule,  better  iron  will  be  obtained  by  purchasing  at  a 
mill  where  only  superior  qualities  are  manufactured,  rather 
than  at  mills  where  all  qualities  are  made. 

In  England  the  brands  are  very  various,  but  are  often  graded 
"Best,"  "Best  Best,"  and  "Treble  Best,"  the  "Best"  being  the 
poorest  of  the  three. 

In  the  United  States  these  brands  are  not  used,  and  the 
usage  varies  with  different  mills.  Some  years  ago  the  brands 
manufactured  were  as  follows  :  — 

i°.  Tank-Iron,  the  lowest  grade,  not  suitable  for  use  in  the 
shell  of  a  boiler,  and  including  all  iron  too  poor  for  this  purpose. 

2°.  C.  No.  I.  — This  used  to  denote  iron  made  by  using  only 
charcoal  in  the  blast-furnace ;  and  it  furnished  a  good  grade  of 
iron,  suitable  to  use  in  those  parts  of  the  shell  of  a  steam-boiler 
which  do  not  come  in  contact  with  the  fire,  and  which  do  not 
have  to  be  flanged. 

3°.  C.  H.  No.  /,  or  charcoal-hammered  No.  i,  which  was 
also  a  charcoal  iron,  but  was  obtained  by  more  working  than 
C.  No.  i,  and  was  hence  a  better  grade  of  iron. 

4°.  Fire-box  Iron,  suitable  to  use  for  sheets  exposed  to  the 
fire. 

5°.  Flange  Iron,  or  such  as  will  flange  without  cracking. 

These  brands  partly  remain  to-day,  but  they  no  longer  bear 
the  same  meanings.  Indeed,  the  name  "  C.  H.  No.  I  "  is,  in 
many  mills,  given  to  iron  in  the  manufacture  of  which  no  char- 
coal has  been  used. 


COMPRESSIVE   STRENGTH  OF    WROUGHT-IRON.         42$ 

There  are  some  mills  that   use   the    following   brands    for 
boiler-plates  ;  viz., — 

Tank- Iron.  Extra  Flange. 

Refined  Iron.  Shell  Fire-Box. 

Shell  Iron.  C.  H.  No.  i  Fire-Box. 

C.  H.  No.  i.  Extra  Flange  Fire-Box. 

Nevertheless,  if  we  buy  any  one  of  these  grades  at  different 
mills,  we  have  no  certainty  of  obtaining  the  same  quality  of 
iron  ;  and,  moreover,  there  are  other  mills  that  use  different 
brands.  "  Refined  iron,"  for  instance,  is  a  term  that  includes  a 
large  number  of  qualities,  beginning  at  much  too  low  a  grade  to 
put  in  a  boiler-shell. 

The  only  sure  way  to  secure  good  iron  is  to  prescribe  the 
tests  it  shall  stand  :  i.e.,  the  tensile  strength,  which  should  be 
over  45000  or  46000  Ibs.  ;  the  limit  of  elasticity,  which  should 
be  as  much  as  27000  or  28000  Ibs. ;  the  contraction  of  area  at 
fracture,  which  should  be  at  least  30  per  cent.  It  should  also 
stand  bending  double  cold,  red-hot,  and  at  a  flanging  heat.  It 
would  be  well  to  test  also  the  soundness  of  the  plates  by 
punching  them. 

BRANDS    OF    BARS    AND    SHAPES. 

For  these,  each  mill  has  its  own  peculiar  brands,  to  which  it 
attaches  its  own  signification.  The  only  way  to  secure  iron  of 
whose  quality  we  are  sure,  is,  therefore,  to  require  that  it  shall 
stand  the  suitable  tests  before  being  accepted. 

COMPRESSIVE    STRENGTH    OF    WROUGHT-IRON. 

In  regard  to  the  compressive  strength  of  wrought-iron,  we 
may  wish  to  study  it  with  reference  to — 

1°.  The  strength  of  wrought-iron  columns; 

2°.  The  strength  of  wrought-iron  beams  ; 

3°.  The  effects  of  a  crushing  force  upon  small  pieces  not 
laterally  supported ; 

4°.  The  effects  of  a  crushing  force  upon  small  pieces  laterally 
supported. 


424  APPLIED   MECHANICS. 

1°.  In  this  case  it -may  be  said  that,  by  reference  to  the 
tests  of  wrought-iron  bridge  columns,  the  compressive  strength 
per  square  inch  of  wrought-iron  in  masses  of  such  sizes  is  given 
by  the  tests  of  the  shorter  lengths  of  such  columns,  i.e.,  by 
those  that  are  short  enough  not  to  acquire,  when  the  maximum 
load  is  just  reached,  a  deflection  sufficient  to  throw  any  appreci- 
ably greater  stress  per  square  inch  on  any  part  of  the  column  in 
consequence  of  the  eccentricity  of  the  load  due  to  the  deflec- 
tion. The  results  thus  obtained  are  naturally  somewhat  lower 
than  we  should  expect  to  obtain  in  smaller  masses. 

2°.  In  this  case  the  evidence  that  there  is,  goes  to  show 
that  the  tensile  and  compressive  strength  of  the  same  iron  are 
nearly  equal.  Some  more  remarks  on  this  subject  will  be  made 
under  "  Transverse  Strength  of  Wrought-iron." 

3°.  If  a  small  cylinder  of  ductile  wrought-iron  is  tested 
without  lateral  support,  and  with  flat  ends,  the  friction  of  the 
ends  against  the  platforms  of  the  testing-machine  comes  in  to 
interfere  with  the  flow  of  the  metal ;  and  if,  besides  this,  the 
ratio  of  length  to  diameter  is  so  small  as  to  prevent  buck- 
ling, then  the  specimen  will  gradually  flatten  out,  and  it 
becomes  impossible  to  find  any  maximum  load,  because  the 
area  of  the  central  part  is  constantly  increasing.  In  case, 
however,  we  call,  at  any  one  time,  the  load  per  square  inch, 
the  load  divided  by  the  largest  section,  at  the  time  the  load  is 
on,  we  shall  find  a  maximum  load  per  square  inch,  in  regard  to 
which  in  the  case  of  mild  steel  there  are  a  number  of  tests  in 
the  Government  reports;  and  from  the  figures  there  given  it 
follows  that  this  maximum  load  per  square  inch  averages  sgme- 
where  near  85000  Ibs. 

4°.  In  this  case  the  crushing  strength  per  square  inch  that 
causes  continuous  flow,  and  also  the  maximum  strength  per 
square  inch,  is  greater  than  that  where  the  specimen  has  no 
lateral  support.  Hence  follows,  that  in  the  case  of  steel  rivets 
it  is  entirely  safe  to  allow  a  bearing  pressure  very  considerably 
in  excess  of  85000  pounds  per  square  inch. 


WROUGHT-IRON   COLUMNS.  42$ 

§  223.  Wrought-Iron  Columns. — Until  a  few  years  ago, 
we  have  had  no  experimental  knowledge  on  this  subject 
beyond  the  experiments  of  Hodgkinson,  which  have  furnished 
the  constants  for  Hodgkinson's,  and  also  for  Gordon's,  formula, 
as  already  given  in  §  209  and  §211. 

These  formulae  have  been  in  very  general  use,  and  it  is  only 
of  late  years  that  we  have  been  able  to  test  their  accuracy  by 
tests  on  full-size  wrought-iron  columns.  The  disagreement  of 
the  formulae  already  referred  to,  with  the  results  of  the  tests,  has 
led  to  the  proposal  of  a  large  number  of  similar  formulae,  each 
having  its  constants  determined  to  suit  a  certain  definite  set  of 
tests,  and  hence  all  these  formulae  thus  proposed  must  be  classed 
as  empirical  formulae,  and  can  only  be  applied  with  safety  within 
the  range  of  the  cases  experimented  upon. 

A  few  of  these  will  now  be  enumerated  ;  and  then  will 
follow  tables  of  the  actual  tests,  which  furifish  the  best  means 
of  determining  the  strength  of  these  columns ;  and  it  would 
appear  that  it  is  these  tables  themselves  which  the  engineer 
would  wish  to  use  in  designing  any  structure. 

On  the  1 5th  of  June,  1881,  Mr.  Clark,  of  the  firm  of  Clark, 
Reeves,  &  Co.,  presented  to  the  American  Society  of  Civil 
Engineers  a  report  of  a  number  of  tests  on  full-size  Phoenix 
columns,  made  for  them  at  the  Watertown  Arsenal,  together 
with  a  comparison  of  the  actual  breaking-weignts  with  those 
which  would  have  been  obtained  by  using  the  common  form  of 
Gordon's  formula  for  wrought-iron, 


where  P  =  breaking-weight  in  Ibs.,  A  =  area  of  section  in 
square  inches,  /  =  length  in  inches,  p  =  least  radius  of  gyra- 
tion in  inches.  The  table  is  as  follows  :  — 


426 


APPLIED   MECHANICS. 


: 

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WROUGHT-IRON    COLUMNS.  427 

The  very  considerable  disagreement  between  the  breaking- 
loads  as  calculated  by  Gordon's  formula,  and  the  actual  break- 
ing-loads, led  a  number  of  people  to  propose  empirical  formulae 
of  one  form  or  another  which  should  represent  this  set  of 
tests,  and  also  Bothers  which  should  represent  some  other  tests 
on  full-size  bridge  columns,  which  "had  been  previously  made 
in  other  places. 

Of  these  I  shall  only  give  those  proposed  by  Mr.  Theodore 
Cooper,  which  are  as  follows :  — 

P  f 

For  square-ended  columns  .     .     .     -j 


For  pin-ended  columns  . 


18000 
/ 


j.Sooo 
And  he  gives,  for  the  values  of/, 

For  Phoenix  columns  .  .  Vr •'•••  .  .  ./=  36000. 
For  American  Company's  columns .  .  .  f  =  30000. 
For  box  and  open  columns f  =  31000. 

He  deduces  these  values  of /from  some  tests  made  in  1875 
by  Mr.  Bouscaren,  combined  with  those,  already  referred  to, 
made  at  the  Watertown  Arsenal.  The  box  and  open  columns 
were  made  of  channel-bars  and  latticing.  The  tables  or  dia- 
grams presented  to  justify  the  formulae  proposed  can  be  found 
in  the  "  Transactions  of  the  American  Society  of  Civil  Engin- 
eers" for  1882. 

Besides  the  above,  there  will  be  given  here  some  formulae 
of  a  similar  form,  derived  by  Prof.  J.  Sondericker,  and  covering 
practically  all  the  sets  of  bridge  columns  tested  on  the  Govern- 
ment machine  through  the  year  1887. 


428  APPLIED   MECHANICS. 

P 

The    formulae    give   approximately   the   values   of  -j,  the 

fi 

breaking-strength  per  square  inch  of  the  columns,  of  the  kinds 
indicated,  when   the    ratios  of   length    to    radius  of   gyration 

are  greater  than  the  number  which   follows  --  inside  of  the 

brackets. 

For  smaller  ratios  of  length  to  radius  of  gyration  we  have  a 
case  of  direct  compression,  and  should  simply  multiply  the 
sectional  area  by  the  compressive-strength  per  square  inch  of 
the  material  in  order  to  obtain  the  breaking-load  of  the  column. 
Moreover,  the  average  compressive-strength  per  square  inch  of 
the  material  for  the  special  columns  tested  is  the  numerator 

P 

of  the  value  of  -j. 

The  numbers  remaining  in  the  formulae  which  have  not 
been  referred  to  here  are  determined  empirically  from  the  sets 
of  tests.  These  formulae  are  purely  empirical,  and  they  are 
given  here  in  preference  to  some  of  the  others,  because,  while 
the  others  deal  only  with  the  earlier  tests,  these  include  the 
entire  range  up  to  date.  The  formulae  are  as  follows  :  — 

For  flat-ended  Phoenix  columns  he^  recommends  Cooper's 
formula. 

Fo?  the  lattice  columns  with  pin-ends,  reported  in  Exec. 
Doc.  12,  47th  Congress,  ist  session,  and  Exec.  Doc.  5,  48th 
Congress,  1st  session,  he  recommends  the  formula 

P  __         34000 
A=        .   II     ' 


I2OOO 


For  the  box  and  solid  web  columns  reported  in  Exec.  Doc.  5, 
48th  Congress,  1st  session,  and  Exec.  Doc.  35,  49th  Congress, 


WROUGHT-IRON  COLUMNS.  429 

ist  session,  taken  together  with  Bouscaren's  results  on  box 
and  on  American  Bridge  Company's  columns,  he  recommends 

P              33000 
For  flat-ends —  = n ^a« 

--80 


1 0000 


For  pin-ends 


5  —         3*000 


6000 


For  other  cases  more  experiments  are  needed  before  satis- 
factory formulae  can  be  devised. 

The  tests  made  at  the  Watertown  Arsenal  will  next  be 
given,  together  with  cuts  showing  the  form  of  the  columns ; 
these  being  taken  from  the  following  government  reports :  — 

Exec.  Doc.  12,  47th  Congress,  ist  session,  House. 
Exec.  Doc.  i,  47th  Congress,  2d  session,  Senate. 
Exec.  Doc.  5,  48th  Congress,  ist  session,  Senate. 
Exec.  Doc.  35,  4Qth  Congress,  ist  session,  Senate. 
Exec.  Doc.  36,  49th  Congress,  ist  session,  Senate. 

The  following  tables  are  taken  from  the  first  of  the  above- 
mentioned  Executive  Documents :  — 


430 


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c'00  OO 

o  K 

srl^ 

,c 
| 

O  "*• 

1 

*i«ii 

»,8. 

<D   C) 

j 

.nX  X 

OQ 

t^.  r^. 

6  d 

1 

""  S 

"rt   C 

ES 

^H 

pi 

WROUGHT-IRON  COLUMNS. 


435 


436 


APPLIED   MECHANICS. 


—]•-•{-  n  i. 


D 


WROUGHT-IRON  COLUMNS. 


437 


The  next  table  taken  from  the  second  Exec.  Document 
mentioned  above  contains  the  results  of  some  compressive  tests 
of  wrought-iron  I-beams  placed  in  the  machine  with  the  ends 
vertical  and  tested  with  flat-ends  ;  also  of  some  tensile  speci- 
mens cut  off  from  two  of  them. 

TESTS   OF   I-BEAMS   BY   COMPRESSION. 


Width 
of 

Thick- 
ness   of 

Total 

.a 
be 

Sectional. 

Ultimate  Strength. 

Length. 

Flange. 

Web. 

Depth. 

I 

Area. 

Actual. 

PerSq.In 

In. 

In. 

In. 

In. 

L  s. 

Sq.  In. 

Lbs. 

Lbs. 

I 

57.06 

5-45 

0.64 

9-00 

228 

14.40 

545100 

37854 

2 

155-45 

4.40 

0.40 

10.52 

443 

IO.26 

207000 

20170 

3 

191.90 

3-56 

0.40 

9.08 

365 

6.85 

85380 

12460 

4 

191.90 

3-59 

0-43 

9.09 

38i 

7.15 

85200 

11916 

5 

119.85 

2.98 

0.28 

6.  ii 

139 

4.18 

IOI2OO 

24210 

6 

180.33 

3-6o 

0.42 

6.96 

303 

6.05 

84650 

13990 

7 

192.04 

3.58 

0-45 

7-94 

355 

6.65 

83400 

12540 

8 

192.90 

3-6o 

0.44 

7.98 

353 

6-59 

92300 

I40IO 

9 

215.88 

4.28 

0.40 

10.52 

56i 

9-30 

I49OOO 

16020 

10 

264.08 

4.49 

0.48 

10.53 

747 

10.19 

II3IOO 

IIIOO 

ii 

264  .  08 

4-43 

0.50 

10.51 

767 

10.46 

107800 

10306 

12 

264.00 

4.90 

0-53 

15.15 

1085 

14.80 

184700 

12400 

13 

263.95 

4.84 

0-53 

14.74 

1081 

14.74 

187000 

12686 

TESTS   OF  SPECIMENS  FROM   NOS.    I    AND   2   BY  TENSION. 


Cut    from 
Flange 
or  Web. 

Width. 
In. 

Depth. 
In. 

Sectional 
Area. 

Sq.  In. 

Ultimate  Strength. 

Contrac- 
tion of 
Area. 

Per  Cent. 

Actual. 
Lbs. 

Per  Sq.  In. 
Lbs. 

r 

Web. 

3-00 

0.65 

1-95 

103300 

52970 

10 

T 

Web. 
Flange. 

3.00 
4.00 

0.50 
0-75 

I-5I 
3-01 

65400 
146400 

43340 
48640 

3-9 
19.6 

I 

Flange. 

4.00 

0.76 

3.02 

147100 

48640 

15.9 

f 

Flange. 

3.OO 

0.51 

i-53 

55400 

36210 

II.  I 

!• 

Web. 

3-00 

0.40 

1.19 

52900 

44640 

I6.5 

43$  APPLIED  MECHANICS. 

Next  will  be  given  the  set  of  tests  which  is  reported  in 
Exec.  Doc.  5,  48th  Congress,  ist  session,  and  Exec.  Doc.  35, 
49th  Congress,  ist  session. 

The  following  is  quoted  from  the  first  of  the  two  : 


"  COMPRESSION    TESTS    OF    WROUGHT-IRON  COLUMNS,   LATTICED,  BOX, 

AND    SOLID    WEB. 

"  This  series  of  tests  comprises  seventy-four  columns,  forty 
of  the  number  having  been  tested,  the  results  of  which  are 
herewith  presented. 

"The  columns  were  made  by  the  Detroit  Bridge  and  Iron 
Company. 

"  The  styles  of  posts  represented  are  those  composed  of — 

"  Channel-bars  with  solid  webs  ; 

"  Channel-bars  and  plates  ; 

"  Plates  and  angles  ; 

"  Channel-bars  latticed,  with  straight  and  swelled  sides  ; 

"  Channel-bars,  latticed  on  one  side,  and  with  continuous 
plate  on  one  side. 

"  All  the  posts  were  tested  with  3^-inch  pins  placed  in  the 
centre  of  gravity  of  cross-section  ;  except  two  posts  of  set  N, 
which  had  the  pins  in  the  centre  of  gravity  of  the  channel- 
bars. 

"  This  gave  an  eccentric  loading  for  these  columns,  on  ac- 
count of  the  continuous  plate  on  one  side  of  the  channel- 
bars. 

"  The  pins  were  used  in  a  vertical  position,  unless  other- 
wise stated  in  the  details  of  the  tests. 

"  In  the  testing-machine  the  posts  occupied  a  horizontal 
position. 

"  They  were  counterweighted  at  the  middle. 

"  Cast-iron  bolsters  for  pin-seats  were  used  between  the  ends 


WROUGHT-IRON  COLUMNS.  439 

of  the  columns  and  the  flat  compression  platforms  of  the  test- 
ing-machine. 

"  The  sectional  areas  were  obtained  from  the  weights  of  the 
channel-bars,  angles,  and  plates,  which  were  weighed  before 
any  holes  were  punched,  calling  the  sectional  area,  in  square 
inches,  one-tenth  the  weight  in  pounds  per  yard  of  the  iron. 

"  Compressions  and  sets  were  measured  within  the  gauged 
length  by  a  screw  micrometer. 

"The  gauged  length  covered  the  middle  portion  of  the 
post,  and  was  taken  along  the  centre  line  of  the  upper  chan- 
nel-bar or  plate,  always  using  a  length  shorter  than  the  dis- 
tance between  the  eye-plates,  to  obtain  gaugings  unaffected  by 
the  concentration  of  the  load  at  those  points. 

"  The  deflections  were  measured  at  the  middle  of  the  post. 
The  pointer,  moving  over  the  face  of  a  dial,  indicated  the 
amount  and  direction  of  the  deflection. 

"  Loads  were  gradually  applied,  measuring  the  compressions 
and  deflections  after  each  increment ;  returning  at  intervals  to 
the  initial  load  to  determine  the  sets. 

"  The  maximum  load  the  column  was  capable  of  sustaining 
was  recorded  as  the  ultimate  strength,  although,  previous  to 
reaching  this  load,  considerable  distortion  may  have  been  pro- 
duced. 

"  Observations  were  made  on  the  behavior  of  the  posts 
after  passing  the  maximum  load,  while  the  pressure  was  fall- 
ing, showing,  in  some  cases,  a  tendency  to  deflect  with  a  sudden 
spring,  accompanied  by  serious  loss  of  strength. 

"  The  slips  of  the  eye-plates  along  the  continuous  plates 
and  channel-bars  during  the  tests  were  measured  for  certain 
posts  in  sets  F,  G,  H,  and  /.  The  measurements  of  slip  were 
taken  in  a  length  of  10  inches  or  20  inches,  one  end  of  the 
micrometer  being  secured  to  the  eye-plate,  and  one  end  to  the 
channel-bar.  The  readings  include  both  the  compression 
movement  of  the  material  and  the  slip  of  the  plates. 


44°  APPLIED  MECHANICS. 

"  Columns  //,  7,  Ly  and  J/ were  provided  with  pin-holes  for 
placing  the  pins  either  parallel  or  perpendicular  to  the  webs  of 
the  channel-bars. 

"  After  the  ultimate  strength  had  been  determined  with  the 
pins  in  their  first  position,  a  supplementary  test  was  made,  if 
the  condition  of  the  column  justified  it,  with  the  pins  at  right 
angles  to  their  former  position  ;  thus  changing  the  moment  Of 
inertia  of  the  cross-section,  taken  about  the  pin  as  an  axis. 

"  The  experiments  with  columns  N  show  how  much  strength 
is  saved  by  employing  pins  in  the  centre  of  gravity  of  the  cross- 
section.  Where  such  was  not  the  case,  the  columns  showed 
the  effect  of  the  eccentric  loading  by  deflections  perpendicular 
to  the  axis  of  the  pins,  from  the  initial  loads,  which  resulted  in 
their  early  failure."  '  *: 


WROUGHT-IRON   COLUMNS. 


441 


TABULATION   OF   EXPERIMENTS   ON   WROUGHT-IRON   COLUMNS 
WITH    3^-INCH   PIN-ENDS. 


No.  of 
Test. 

Style  of  Column. 

Length, 
Centre 
to 
Centre 
of  Pins. 

In. 

Sec- 
tional 
Area. 

Sq.  In. 

Ultimate 
Strength. 

Manner  of  Failure. 

Total, 
Lbs. 

Lbs. 
per 
Sq.  In. 

Set!  A 

752 

1  T  1 

126.20 

9-831 

297100 

30220 

Deflected    perpendicular 
to  axis  of  pins. 

757 

120.07 

10.199 

320000 

31380 

Sheared     rivets    in    eye- 

lf\ 

t  1 

plates. 

755 
756 

d 

CO 

b* 

180.00 
180.00 

9-977 
9-977 

251000 

210000 

25160 
21050 

Deflected    perpendicular 
to  axis  of  pins. 
Do.             do. 

753 

V 

240.00 

9-732 

I88600 

19380 

Do.             do. 

754 

3 

* 

240  .  10 

9.762 

158300 

16220 

Do.             do. 

Set',D, 

1642 

I 

240.00 
240.00 

16.077 
16.281 

425000 
367000 

26430 
22540 

Deflected    perpendicular 
to  axis  of  pins. 
Do.            do. 

1 

r  T 

1646 

1    'I 

320.00 

16.179 

318800 

19700 

Do.             do. 

1647 

320.10 

16.141 

283600 

I757° 

Do.             do. 

1    Set 

*-  —  » 

!*. 

1654 

IC/* 

.J 

oo 

320.00 
320.00 

17-898 
19.417 

474000 
491000 

26480 
25290 

Deflected    perpendicular 
to  axis  of  pins. 
Do.            do. 

<-£ 

1 

Set  F. 

1645 

*=          I    t 

Tin 

3I9-95 

16.168 

453000 

28020 

Deflected  parallel  to  axis 

J'1 

ft      •      ~*OO 

of  pins. 

1650 

F* 

[L 

320.00 

16.267 

454000 

27910 

Deflected    perpendicular 
to  axis  of  pins. 

*? 

442 


APPLIED  MECHANICS. 


TABULATION   OF   EXPERIMENTS   ON  WROUGHT-IRON  COLUMNS 
WITH    3^-INCH   PIN-ENDS. 


Length, 
Centre 

Sec- 

Ultimate 
Strength. 

No.  of 
Test. 

Style  of  Column. 

to 
Centre 

tional 
Area. 

Lbs. 

Manner  of  Failure. 

of  Pins. 

Total, 
Lbs. 

per 

In. 

Sq.  In. 

Sq.  In. 

Set-G. 

1651 

,,T 

320.00 

20.954 

540000 

25770 

Deflected      in     diagonal 

'           „          8=Y 

direction. 

1652 

320.10 

20.613 

535000 

25950 

Sheared    rivets    in    eye- 

| 

>"      J 

_ 

plates. 

746 
747 
748 

pq 

SetH. 

d;: 

R 

159-20 
159-27 
239.60 

7.628 
8.056 
7.621 

258700 
294700 
260000 

36580 
34120 

Deflected    perpendicular 
to  axis  of  pins. 
Do.             do. 

Do.             do. 

749 

I 

239.60 

7  621 

254600 

33410 

Deflected      in     diagonal 

1648 

cr= 

_    1 

3J9-90 

7.705 

243600 

U  tl  Cl*UUIl, 

31610  (Deflected  parallel  to  ?xis 

of  pins. 

1649 

1 

319-85 

7^73 

229200 

29870 

Deflected      in     diagonal 

direction. 

740 

Set  I. 

i59-9o 

7.<H5 

262500 

34340 

Deflected    perpendicular 
to  axis  of  pins. 

74i 

(swejled.)__ 

i59-9o 

7.624 

255650 

33530 

Do.             do. 

739 

239.70 

7-5^7 

251000 

33390 

Deflected  parallel  to  axis 

1 

J 

of  pins. 

750 

5  (-8    —  =* 

* 

239.70 

7-531 

259000 

34390 

Deflected     perpendicular 

I 

i 

to  axis  ol  pins. 

1643 

1 

I 

319-80 

7.691 

237200 

30840 

Deflected  parallel  to  axis 

•  1    II    •   •!        !!•••• 

of  pins. 

1644 

319.92 

7.702 

237000 

30770 

Deflected     in      diagonal 

L 

direction. 

1640 
1641 

1634 

— 

4! 

r! 

199.84 
200.00 
300.00 

11.944 
I2.3O2 
12.148 

403000 
426500 
408000 

33740 
34670 
3363° 

Deflected    perpendicular 
to  axis  of  pins. 
Deflected      in      diagonal 
direction. 
Deflected     perpendicular 

•*—  ' 

to  axis  of  pins. 

1635 

i 

_i 

300.00 

12.175 

395000 

32440 

Do.             do. 

i 

WROUGHT-IRON  COLUMNS. 


443 


TABULATION  OF    EXPERIMENTS   ON  WROUGHT-IRON   COLUMNS 
WITH    3J-INCH   PIN-ENDS. 


Length, 
Centre 

Sec- 

Ultimate 
Strength. 

No.  of 
Test. 

Style  of  Column. 

to 
Centre 

tional 
Area. 

Total, 

Lbs. 

Manner  of  Failure. 

of  Pins. 

Lbs. 

per 

In. 

Sq.  In. 

Sq.  In. 

SeitM. 

(swelled.) 

1638 
1639 

J 

199-25 
199.50 

12.366 
12.659 

385000 
405000 

31990 

Deflected    perpendicular 
to  axis  of  pins. 
Do.             do. 

<      "'            '   g            —  ^ 

§ 

1636 

I 

300.20 

11.920 

391400 

32830 

Deflected     in     diagonal 
direction. 

1637 

sa  —  J  

_1 

300.15 

11.932 

390700 

32740 

Do.             do. 

i 

1630 

SejtN. 

^r-Sfr-tf-- 

I 

300.00 

17.622 

461500 

26190 

Deflected    perpendicular 
to  axis  of  pins. 

1631 

•e 

300.00 

17.231 

485000 

28150 

Do.             do. 

1632 

|p£.  ioj'  !  — 

.1 

300.00 

I7-570 

306000 

17420 

Do.            do. 

1633 

J     •    J    : 

i     -I-  £      g 

"JR 

300.00 

17.721 

307000 

17270 

Do.            do. 

Ij          S     (2 

1 

•sl    = 

1        =• 

1 

The  remainder  of  the  tests  of  this  series  of  seventy-four 
columns  is  reported  in  Exec.  Doc.  35,  49th  Congress,  1st  ses- 
sion. 

The  only  portion  of  the  description  that  it  is  worth  while 
to  quote  is  the  following,  as  the  tests  were  made  in  a  similar 
way  to  what  has  been  already  described : 

"  Sixteen  posts  were  tested  with  flat  ends ;  eighteen  were 
tested  with  3^inch  pin-ends. 


444 


APPLIED  MECHANICS. 


"  The  pins  were  placed  in  the  centre  of  gravity  of  cross- 
section,  except  two  posts  of  set  K,  which  had  the  pins  in  the 
centre  of  gravity  of  the.  channel-bars,  giving  an  eccentric  bear- 
ing to  these  columns,  on  account  of  the  continuous  plate  on 
one  side  of  the  channel-bars." 


TABULATION  OF  EXPERIMENTS   ON  WROUGHT-IRON   COLUMNS 
WITH   FLAT   ENDS. 


Ultimate 

Total 

Sec- 

Strength. 

No.  of 
Test. 

Style  of  Column. 

Length. 

tional 
Area. 

Total, 

Per 

Number  of  Failure. 

T  HQ 

Sq.  In., 

Ft.    In. 

Sq.  In. 

JL«DS. 

Lbs. 

377 

SetB. 

I 

ff 

10     7.90 

12.08 

383200 

31722 

Buckling-plate    D    be- 
tween the  riveting. 

"4 

378 

<-  6'-^—  > 

10     7.90 

11.  ii 

372000 

33564 

Buckling-plates. 

1 

379 

SetE. 

•*••  4 

—  8'L~t* 

BSi 

13  11.80 
13  11.80 

17.01 

17.80 

594500 
633600 

3495° 
35595 

Buckling  -  plates      be- 
tween the  riveting. 
Triple  flexure. 

346 

13  11.9 

15-74 

517000 

32846 

Buckling-plates. 

— 

A  t  II  — 

Tig 

347 

.        ^jft  tL.sx'' 

13  11-65 

15.84 

555200 

35050 

Do.            do. 

Set  P.. 

t^8"     ic  *r% 

342 

L 

20      7.63 

15.68 

517500 

33003 

Deflecting  upward. 

344 

^K." 

2O      7.80 

15-56 

536900 

34505 

Buckling-plates. 

348 

13  "-75 

21.02 

708000 

33682 

Buckling-plates. 

349 

1  *•  4 

Beta]»V     S 

< 

13  "-75 

20      7.60 

21.46 
21.20 

709500 

7OOOOO 

33061 
33019 

Triple  flexure. 
Deflecting  upward. 

343 

[\*~  6.90  -£• 

•^ 

20      7.63 

21.49 

729450 

33943 

Deflecting  downward. 

*3? 

WROUGHT-IRON  COLUMNS. 


445 


TABULATION   OF   EXPERIMENTS   ON  WROUGHT-IRON  COLUMNS 
WITH   FLAT   ENDS. 


Style  of  Column. 

Ultimate 

Total 

Sec- 

Strength. 

No.  of 
Test. 

Length. 

tional 
Area. 

Manner  of  Failure. 

Total, 

Per  Sq. 

Ft.    In. 

Sq.  In. 

Lbs. 

In.,  Ibs. 

= 

Ka'We/f 

=  t 

339 

,t  1 

20      7.94 

12.64 

412900 

32666 

Deflecting  upward. 

SetK. 

I 

340 

4-  8-  —  ¥ 

I 

20      7.94 

12.74 

431400 

33862 

Do.             do. 

"Latticed 

-=W|r 

•== 

He'W" 

"] 

337 

338 

SetN. 

<-    8"-* 

Latticed 

u 

25      7-75 

25    7-88 

16.99 
17.40 

582400 
580000 

34279 
33333 

Deflecting     downward 
and  sideways. 
Deflecting     diagonally 
channel  B  and  lattic- 
ing on    the    concave 
side. 

TABULATION   OF   EXPERIMENTS   ON  WROUGHT-IRON   COLUMNS 
WITH    3-J-INCH   PIN-ENDS. 


Length,! 

Ultimate 

Centre 

Sec- 

Strength. 

No.  of 
Test. 

Style  of  Column. 

to 
Centre 
of  Pins. 

tional 
Area. 

Manner  of  Failure. 

Total, 

Per  Sq. 

Ft.    In. 

Sq.  In. 

Lbs. 

In.,  Ibs. 

368 
367 
356 

i! 

* 

15      o.i 
15      o.o 
20      o.o 
20      o.o 

11.42 
11.42 
11.42 
11.31 

379200 
369200 
342000 
330100 

33205 

32329 

29947 
29186 

Hor.  deflection  perpen- 
dic.  to  plane  of  pins. 
Hor.  deflection. 

Do.             do. 
Do.             do. 

<  ,  °t 

s<«.^6,  r 

n\ 

37i 

9     "-9 

9.14 

286100 

31302 

Buckling  -  plates      be- 

n 

tween  rivets. 

372 

.        Ki 

10         0.0 

10.07 

319200 

31698 

Do.             do. 

370 

i  " 

15      o.o 

9.21 

291500 

3^50 

Hor.  deflec.  and  buck- 

369 

Set(2J*5.W''l> 

L 

15       o.o 

9-44 

290000 

30720 

ling  between  rivets. 
Do.             do. 

354 

20      o.o 

9.24 

267500 

28950 

Triple  flexure. 

365 

20         0.0 

9-36 

279700 

29879 

Hor.  deflection. 

446 


APPLIED   MECHANICS. 


TABULATION   OF   EXPERIMENTS   ON   WROUGHT-IRON   COLUMNS 
WITH    3|-INCH    PIN-ENDS. 


Style  of  Column. 

Length, 
Centre 

Sec- 

Ultimate 
Strength. 

No.  of 
Test. 

to 
Centre 

tional 
Area. 

Total, 

Per 

Manner  of  Failure. 

of  Pins. 

Lbs. 

Sq.  In., 

Ft.    In. 

Sq.  In. 

Lbs. 

3 

]  ;*  # 

360 
361 

13    4-i3 
13    4.00 

J5-34 
15.40 

475000 
485000 

30965 
3M94 

Deflecting    upward    in 
plane  of  pins. 
Hor.  deflection  perpen- 
dicular   to    plane    of 

oins. 

1 

'                | 

358 
359 

Set  E. 

—  s" 

3 

,*." 

20      0.0 

20    o.o 

17.77 
17.22 

570000 
5554oo 

32077 
32253 

Hor.  deflection  perpen- 
dicular   to    plane    of 
pins. 
Do.             do. 

, 

*i 

j  > 

350. 

i 

20      0.25 

12.48 

202700 

16242 

Hor.     deflection,     con- 

(•STI 

^ 

i     cave  on  lattice  side. 

35i 

SctK. 

II 

00 

I 

20      0.00 

10.84 

208200 

19207 

Do,            .do. 

352 

j 

-    1 

i 

20      0  .  25 

12.65 

' 

350000 

37668             Do.             do. 

*e« 

«*-(  1 

. 

353 

"1 

~    ^ 

20      0.25 

12.76 

390400 

30596    Hor.  deflection  perpen- 
j     dicular    to    plane    of 

"1 

; 

pins,    convex  on  lat- 

tice side. 

Besides  the  above,  there  are  four  tests  of  lattice  column^ 
reported  in  Exec.  Doc.  36,  49th  Congress,  1st  session,  but  as 
these  columns  were  rather  poorly  constructed  and  form  rather 
special  cases  they  will  not  be  quoted  here. 

In  determining  the  strength  of  a  bridge  column  made  of 
channel-bars  and  latticing,  these  results  of  tests  on  full-size 
columns  furnish  us  the  best  data  upon  which  to  base  our  con- 
clusions ;  and  the  formulae  proposed  by  the  different  persons 


TRANSVERSE  STRENGTH  OF  WROUGHT-IRON.  447 

who  have  deduced  formulae  to  cover  special  sets  of  values  may 
be  advantageously  used  for  cases  intermediate  between  those 
that  have  been  experimented  upon. 

§  224.  Transverse  Strength  of  Wrought-Iron  and 
Steel.  —  Wrought-iron  owes  its  extensive  introduction  into 
construction  as  much  or  more  to  the  efforts  of  Sir  William 
Fairbairn  than  to  any  one  else ;  and  while  he  was  furnishing 
the  means  to  Eaton  Hodgkinson  to  make  extensive  experiments 
on  cast-iron  columns,  and  while  he  made  experiments  himself 
on  cast-iron  beams,  which  were  in  use  at  that  time,  he  also 
carried  on  a  large  number  of  tests  on  beams  built  of  wrought- 
iron,  more  especially  those  of  tubular  form,  and  those  having 
an  I  or  a  T  section,  and  made  of  pieces  riveted  together.  In 
his  book  on  the  "Application  of  Cast  and  Wrought  Iron  to 
Building  Purposes "  he  gives  an  account  of  a  large  number 
of  these  experiments,  including  those  made  for  the  purpose  of 
designing  the  Britannia  and  Conway  tubular  bridges,  a  fuller 
account  of  which  will  be  found  in  his  book  entitled  "An  Ac- 
count of  the  Construction  of  the  Britannia  and  Conway  Tubular 
Bridges."  In  the  first-named  treatise  he  urges  very  strongly 
the  use  of  wrought-iron,  instead  of  cast-iron,  to  bear  a  trans- 
verse load. 

Fairbairn  tested  a  number  of  wrought-iron  built-up  beams 
with  different  thicknesses  of  upper  and  lower  flange.  The  first 
results  he  obtained  showed,  that,  unless  the  upper  plates  were 
made  very  much  thicker  than  the  lower,  the  beam  would  invari- 
ably give  way  by  crippling  at  the  top. 

At  first  he  concluded  that  the  tensile  strength  of  wrought- 
iron  is  greater  than  its  compressive  strength,  and  it  has  often 
been  so  regarded. 

Subsequent  experiments  made  by  him,  however,  showed, 
that,  if  the  iron  of  the  upper  flange  were  distributed  in  the  form 
of  cells,  the  areas  of  the  cross-sections  required  in  order  to 
render  the  beam  equally  liable  to  give  way  by  tearing  or  by 


448  APPLIED  MECHANICS. 

crushing,  were  as  12  to  n,  or  nearly  equal;  thus  tending  to 
show  that  the  tensile  and  compressive  strengths  of  wrought- 
iron  are  nearly  equal,  and  that  the  reason  of  the  crippling  in 
the  first  experiments  was,  that  the  iron  between  the  rivets 
acted  like  a  column,  and  bent,  instead  of  bringing  into  play  the 
entire  compressive  strength  of  the  iron. 

Experiments  made  on  small  round  or  rectangular  bars  show 
a  modulus  of  rupture  very  considerably  in  excess  of  the  tensile 
or  compressive  strength,  the  explanation  being,  just  as  was 

explained  in  the  case  of  cast-iron,  that  the  formula  M  —  f- 

assumes  Hooke's  law  to  hold,  and  that  this  is  not  true  near  the 
breaking-point. 

As  to  experiments  on  large  beams,  besides  those  of  Fair- 
bairn,  already  referred  to,  we  have :  — 

i°.  Some  tests  made  by  Mr.  William  Sooy  Smith  and  by 
Col.  Laidley  at  the  Watertown  Arsenal. 

2°.  Some  tests  made  in  Holland  on  iron  and  steel  beams, 
an  account  of  which  is  given  in  the  Proceedings  of  the  British 
Institute  of  Civil  Engineers  for  1886,  vol.  Ixxxiv.  p.  412  et  seq. 

3°.  Some  tests  made  in  the  laboratory  of  Applied  Mechanics 
of  the  Massachusetts  Institute  of  Technology,  on  iron  and 
steel  I-beams. 

4°.  Tests  made  by  the  different  iron  companies  upon  beams 
of  their  own  manufacture,  and  recorded  in  their  respective 
hand-books. 

Mr.  Smith's  tests  are  recorded  in  Executive  Document  23, 
46th  Congress,  second  session,  and,  as  it  does  not  seem  to  the 
writer  that  any  conclusions  can  be  drawn  from  this  series  of 
tests,  it  will  not  be  quoted  here. 

In  the  84th  volume  of  the  Proceedings  of  the  British 
Institution  of  Civil  Engineers  is  given  an  account  of  two  sets 
of  tests  made  in  Holland  in  1877,  upon  such  wrought-iron  and 
steel  girders  as  were  in  use  there  at  that  time. 


TRANSVERSE   STRENGTH  OF    WROUGHT-IRON.         449 

In  the  first  series,  seven  riveted  steel  girders  were  tested, 
giving  for  moduli  of  rupture  in  pounds  per  square  inch  the 
following  values  respectively,  viz.:  35560,  34140,  53/60,  54050, 
62580,  58310,  and  51200;  whereas  the  tensile  strength  of  the 
steel  varied  from  85340  to  106670  pounds  per  square  inch. 

After  obtaining  these  results,  which  were  decidedly  disap- 
pointing to  the  steel  makers,  they  decided  to  make  a  more 
extended  series  of  tests  upon  steel  and  wrought-iron  girders. 
The  girders  tested  in  this  second  series  were,  according  to  the 
paper  referred  to,  the  following,  viz. :  three  wrought-iron,  three 
hard  steel,  three  soft  steel,  and  twenty  of  the  longitudinal 
girders  which  had  been  intended  for  the  bridge  across  the 
river  Waal  at  Nymegen,  but  which  had  been  rejected. 

The  requirements  prescribed  for  the  hard  steel  were  a  mini- 
mum tensile  strength  of  106670  pounds  per  square  inch  and  an 
ultimate  elongation  not  less  than  14  per  cent.  For  soft  steel 
they  were  a  maximum  tensile  strength  of  71120  pounds  per 
square  inch,  and  an  ultimate  elongation  of  25  per  cent. 

The  modulus  of  rupture  obtained  from  the  tests  of  the 
three  wrought-iron  girders  was  55470  pounds  per  square  inch, 
while  the  tensile  strength  of  the  wrought-iron  varied  from 
54050  to  56890  pounds  per  square  inch,  and  their  contraction 
of  area  at  fracture  from  II  per  cent  to  35  per  cent. 

There  had  been  considerable  difficulty  encountered  in  pro- 
curing steel  which  should  comply  with  the  requirements,  in  the 
case  of  the  three  hard  steel  girders  ;  hence  these  three  represent 
a  much  better  quality  of  steel  than  that  which  was  in  use  for 
girders  at  that  time. 

The  moduli  of  rupture  in  pounds  per  square  inch  which 
were  obtained  from  these  three  tests  were  respectively.  96720, 
83920,  and  96720,  the  tensile  strength  of  the  steel  varying  from 
115210  to  122320  pounds  per  square  inch,  and  the  contraction 
of  area  at  fracture  from  24  per  cent  to  36  per  cent.  The 
modulus  of  rupture  obtained  from  the  three  soft  steel  girders 


45°  APPLIED   MECHANICS. 

was  49780  pounds  per  square  inch,  while  the  tensile  strength  of 
the  soft  steel  varied  from  59/40  to  69690  pounds  per  square 
inch,  and  the  contraction  of  area  at  fracture  from  42  per  cent 
to  50  per  cent.  Of  the  twenty  old  girders,  two  were  bolted,  and 
eighteen  riveted.  The  former  gave  moduli  of  rupture  equal  to 
51210  and  71120  pounds  per  square  inch  respectively. 

Of  the  other  eighteen,  seven  gave  moduli  of  rupture  below 
56890  pounds  per  square  inch,  ten  between  this  and  71120, 
while  one  gave  83920. 

The  tensile  strength  of  the  material  varied  from  55470  to 
122320  pounds  per  square  inch,  and  the  contraction  of  area  at 
fracture  from  1 1  per  cent  to  49  per  cent. 

Tests  were  made  to  determine  whether  annealing  different 
parts  of  the  steel  girders  before  riveting  would  be  an  advan- 
tage, but  the  tests  gave  very  unsatisfactory  results. 

It  is  plain  that  of  the  girders  tested  in  these  two  series  of 
tests  the  only  ones  that  according  to  our  modern  standards 
would  be  considered  suitable  for  use  are  the  three  wrought- 
iron*  and  the  three  soft-steel  girders,  inasmuch  as  the  steel  in 
all  the  other  cases  was  too  hard  and  too  variable  in  its  proper- 
ties to  be  suitable  for  use  in  construction. 

The  following  table  gives  the  results  that  have  been 
obtained  in  the  tests  that  have  been  made  upon  wrought-iron 
and  steel  I-beams  in  the  laboratory  of  Applied  Mechanics  of 
the  Massachusetts  Institute  of  Technology.  This  table  will 
give  a  fair  idea  of  the  strength  and  elasticity  of  such  beams  as 
are  in  use  to-day.  Moreover,  it  is  pretty  well  recognized  to-day 
that  for  use  in  construction  we  need  a  soft  and  ductile  steel ; 
and  hence  that  hard-steel  beams  would  be  unsafe  on  account  of 
their  lack  of  ductility. 


TRANSVERSE  STRENGTH  OF  WROUGHT-1RON.          45  I 


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45 2  APPLIED   MECHANICS. 

§  225.  Steel.  —  Steel  is  usually  defined  to  be  a  compound 
of  iron  with  a  small  percentage  of  carbon,  this  percentage 
varying  from  a  very  minute  quantity  up  to  one  and  a  half,  or  at 
most  two,  per  cent.  Steel  is  made  in  one  of  the  three  follow- 
ing ways  ;  viz.,  — 

i°.  By  adding  carbon  to  wrought-iron. 

2°.  By  removing  carbon  from  cast-iron. 

3°.  By  melting  together  cast  and  wrought  iron  in  suitable 
proportions. 

These  three  processes  are,  as  a  rule,  represented  respectively 
by  the  following  three  kinds  of  steel :  — 

i°.  Crucible  steel, 

2°.  Bessemer  steel, 

3°.  Open-hearth,  or  Siemens's  Martin  steel ; 
these  being  the  three  chief  kinds  that  are  made  on  the  com- 
mercial scale. 

Crucible  Steel.  —  This  should  always  be,  and  is  by  good 
makers,  made  by  re-melting  blister  steel  in  crucibles  ;  the  lat- 
ter being  made  by  the  cementation  process,  in  which  bars  of 
wrought-iron  'are  heated  in  contact  with  charcoal  until  they 
have  absorbed  the  necessary  amount  of  carbon. 

Crucible  steel  is  used  for  the  finest  cutlery,  tools,  etc.,  and 
wherever  a  very  pure  and  homogeneous  quality  of  steel  is 
required. 

Bessemer  Steel  is  made  by  decarbonizing  cast-iron  by  forcing 
a  powerful  blast  through  a  mass  of  melted  cast-iron,  thus  re- 
moving the  greater  part  of  its  carbon,  and  then  adding  a  small 
quantity  of  some  very  pure  cast-iron  which  is  rich  in  carbon,  thus 
bringing  up  the  percentage  of  carbon  to  the  required  amount. 

Open-hearth  Steel  is  made  by  fusing  a  charge  consisting  of 
the  suitable  proportions  of  cast-iron  with  wrought-iron  scrap, 
or  with  Bessemer  steel  scrap. 

Bessemer  and  open-hearth  steel  contain  more  impurities 
than  crucible  steel ;  but  they  are  very  much  cheaper,  and  are 


STEEL.  453 


just  as  suitable  for  many  purposes.  It  is  only  in  consequence 
of  their  introduction  that  steel  can  be  extensively  used  on  the 
large  scale,  as  crucible  steel  would  be  too  expensive  for  many 
purposes. 

Steel  is  also  made  by  puddling  and  by  other  processes. 

Steel,  unlike  wrought-iron,  is  fusible  ;  unlike  cast-iron,  it  can 
be  forged ;  and,  with  the  exception  of  the  higher  grades,  it 
can  be  welded  by  heating  and  hammering,  the  welding  of  high- 
grade  steel  in  large  masses  being  a  very  uncertain  operation, 
though  small  masses  can  be  welded  by  taking  proper  care. 

The  special  characteristic,  however,  is,  that,  with  the  ex- 
ception of  the  lowest  grades,  when  raised  to  a  red  heat  and 
suddenly  cooled,  it  becomes  hard  and  brittle,  and  that,  by  sub- 
sequent heating  and  slow  cooling,  the  hardness  may  be  reduced 
to  any  desired  degree.  The  first  process  is  called  hardening, 
and  the  second  tempering. 

Pure  wrought-iron  cannot  be  hardened  by  this  means ;  and 
cast-iron  can  be  hardened,  but  cannot  be  tempered. 

Case  Hardening  is  a  process  by  which  the  outer  coating  of 
wrought-iron  is  turned  into  steel  by  heating  it  to  a  red  heat  in 
contact  with  bone-dust  or  some  animal  matter.  This  process 
gives  the  iron  a  hard  surface  combined  with  toughness. 

Whereas  the  hardening  element  of  steel  should  be  only 
carbon,  and  whereas  other  substances  should  be  absent  as  far 
as  possible  in  the  best  steel,  nevertheless  phosphorus,  silicon, 
and  manganese,  when  present  in  small  quantities,  all  have  a 
hardening  effect ;  and  all  these  ingredients,  and  often  sulphur, 
are  generally  found  in  Bessemer  and  open-hearth  steel,  sul- 
phur, silicon,  and  phosphorus  coming  from  the  ore,  the  fuel, 
and  the  flux,  and  manganese  being  necessarily  added,  partly 
to  counteract  the  effect  of  sulphur,  partly,  by  its  affinity  for 
oxygen,  to  absorb  any  oxygen  in  the  interior  of  the  mass,  and 
thus  decrease  the  porosity,  and  partly  to  enable  the  steel  to  be 
welded  by  preventing  the  rapid  oxidation  of  the  surfaces  at  a 
high  heat. 


454  APPLIED   MECHANICS. 

When  Bessemer  steel  was  first  introduced,  and  for  some 
time  thereafter,  it  was  chiefly  used  for  rails.  It  is  only  since  a 
more  recent  date  that  Bessemer  and  open-hearth  steel  have 
been  made  sufficiently  homogeneous  and  reliable  for  use  in 
construction  generally ;  but  at  the  present  day  it  is  displacing 
wrought-iron  for  very  many  purposes,  as  being  more  reliable, 
and  it  is  likely  to  displace  it  even  more. 

In  the  construction  of  boilers,  bridges,  trusses,  beams,  etc., 
it  will  not  do  to  use  the  higher  grades  of  steel,  but  only  the 
milder  and  more  ductile  kinds  :  thus  steel  with  a  tensile  strength 
of  more  than  80000  or  90000  Ibs.  per  square  inch  is  generally  too 
hard  to  use  in  construction  ;  and  for  steam-boilers,  if  its  strength 
exceed  about  60000  Ibs.,  it  is  liable  to  be  too  hard.  It  should 
also  show  a  large  percentage  of  contraction  of  area,  as  30  per 
cent  or  upwards.  Such  steel  contains  but  little  carbon,  gen- 
erally not  more  than  one-half  of  one  per  cent. 

Welding  by  heating  and  hammering,  i.e.,  not  electro-weld- 
ing, is  a  much  more  difficult  operation  in  steel  than  in  iron,  as 
(i°)  there  is  always  danger  of  overheating,  and  (2°)  the  metal 
does  not  unite  as  readily  at  a  welding-heat ;  hence,  in  high 
grades  of  steel,  welding  is  almost  an  impossibility,  especially 
with  large  masses. 

The  injury  done  to  steel  plates  by  punching  is  greater  than 
that  done  to  iron  plates  :  this  injury  can,  however,  be  removed 
by  annealing.  Steel  requires  greater  care  in  working  it  than 
iron,  whether  in  punching,  flanging,  riveting,  or  other  methods 
of  working ;  otherwise  it  may,  if  over-heated,  burn,  or  receive 
other  injury  from  careless  workmanship; 

In  regard  to  the  term  "temper,"  it  should  be  observed,  that 
by  the  steel-maker  it  is  used  to  denote  the  percentage  of  carbon 
in  the  steel,  a  higher  temper  corresponding  to  a  higher  per- 
centage of  carbon.  On  the  other  hand,  the  term  "temper"  in 
common  parlance  refers  to  the  degree  of  hardness  as  deter- 
mined by  tempering  the  steel. 


TENSILE  AND  COMPRESSIVE  STRENGTH  OF  STEEL.    455 

The  brands  of  steel  are  determined  by  each  maker  for  him- 
self, there  being  no  uniformity  in  this  regard. 

The  chemical  composition  of  steel  is  one  important  element 
in  its  resisting  properties  ;  but,  on  the  other  hand,  the  mode  of 
working  also  has  a  great  influence  on  the  quality. 

The  only  means  of  securing  good  steel  is,  to  prescribe  the 
tests  which  it  shall  stand,  and  to  reject  all  that  does  not  fulfil 
the  requirements. 

Thus,  good  boiler-plate  should  have  an  ultimate  strength 
of  55000  to  60000  Ibs.  per  square  inch,  a  limit  of  elasticity  of 
about  30000  pounds,  a  contraction  of  area  at  fracture  of  about 
30  percent.  It  should  not  crack  on  (i°)  being  bent  double  cold, 
(2°)  at  a  red  heat,  (3°)  at  a  flanging-heat,  and  it  should  suffer 
but  little  injury  by  punching. 

For  other  purposes,  as  in  trusses,  etc.,  it  should  be  able  to 
stand,  without  injury,  the  trials  to  which  it  has  to  be  subjected 
in  construction,  as  bending,  punching,  riveting,  etc.  An  ac- 
count of  the  manner  of  applying  such  tests  to  angle-irons, 
I-beams,  etc.,  will  be  found  in  "  Use  of  Steel  for  Constructive 
Purposes,"  by  J.  Barba. 

§  226.  Tensile  and  Compressive  Strength  of  Steel. — 
No  references  will  be  given  to  any  of  the  very  early  tests  upon 
the  strength  of  steel,  as  the  records  of  the  kinds  of  steel  used 
in  the  tests  and  of  its  mode  of  manufacture  are  very  imperfect. 
By  way  of  references  to  the  literature  on  tests  of  steel,  it  may 
be  said  that  the  greater  part  of  those  given  under  the  head  of 
wrought-iron  contain  also  experiments  on  steel.  Besides  these 
may  be  mentioned  the  following  :  — 

W.  E.  Woodbridge:  Report  on  the  Mechanical  Properties  of  Steel, 
Chiefly  with  Reference  to  Gun  Construction  on  the  Wood- 
bridge  System. 

J.  Barba:  The  Use  of  Steel  for  Constructive  Purposes. 

Tests  made  for  St.  Louis  Arch:  Woodward's  History  of  the  St.  Louis 
Arch. 


45 6  APPLIED  MECHANICS, 

Tests  of  Steel  for  the  Brooklyn  Bridge:  Roebling's  Report  of  the 
Brooklyn  Bridge. 

Charles  B.  Dudley:  Franklin  Institute  Journal,  1881. 

A.  F.  Hill:  Proceedings  of  the  Engineers'  Society  of  Western  Penn- 
sylvania for  1880. 

A.  F.  Hill:  Proceedings  of  Society  of  Arts  of  the  Massachusetts 
Institute  of  Technology,  for  1882-3. 

The  strength,  elasticity,  and  other  properties  of  steel  depend, 
of  course,  upon  its  chemical  composition,  upon  its  mode  of 
manufacture,  and  upon  the  treatment  it  receives  while  it  is 
being  transformed  into  the  finished  product. 

The  engineer  should  be  very  careful  to  prescribe  such  phys- 
ical tests  for  the  steel  which  he  is  to  use  as  will  insure  a  suit- 
able material,  and  among  these  physical  tests  should  be  included 
such  as  will  prove  that  it  will  not  be  injured  by  the  proper 
working  to  which  it  must  necessarily  be  subjected  in  the  manu- 
facture of  the  articles  that  are  to  be  made  of  it ;  but  the 
engineer  should  leave  to  the  steel  manufacturer  to  determine 
how  to  produce  a  steel  that  will  fulfil  the  requirements  pre- 
scribed ;  and  he  should  then  see  to  it  that  it  does  not  receive 
improper  or  careless  treatment  during  its  manufacture  into  the 
finished  product. 

Nevertheless  it  is  desirable  that  the  engineer  should  have 
some  knowledge  of  the  effect  of  the  different  ingredients  upon 
the  strength,  ductility,  etc.,  of  steel. 

While  quite  a  number  of  tests  have  been  made  with  this 
object  in  view,  and  quite  a  number  of  empirical  formulae  have 
been  deduced,  nevertheless  the  tests  have  not,  in  the  writer's 
opinion,  been  sufficiently  full  to  warrant  the  deduction  of  the 
formulae  from  them. 

Thus  several  sets  of  tests  have  been  made  to  determine  how 
the  percentage  of  carbon  affects  the  strength  of  the  steel,  but 
in  none  of  them  have  the  proportions  of  the  other  ingredients 
been  kept  constant ;  hence  we  are  not  warranted  in  drawing  as 


TENSILE  AND  COMPRESSIVE  STRENGTH  OF  STEEL.    457 

full  conclusions  as  we  might,  could  this  condition  have  been 
fulfilled. 

With  the  view  of  presenting  some  information  upon  this 
subject  three  sets  of  tests  will  be  quoted  here. 

The  first  is  a  series,  of  tests  made  by  Bauschinger,  and 
recorded  in  the  third  Heft  of  the  Mittheilungen.  The  steel 
used  was  Bessemer  steel  made  at  Ternitz,  and  for  these  tests 
twelve  charges  were  made,  which  only  differed  in  the  amount 
of  Spiegel  iron  added  ;  and  the  percentage  of  carbon  was  deter- 
mined by  chemical  analysis  of  the  steel. 

The  following  table  gives  his  results  upon  the  tensile,  com- 
pressive,  shearing,  and  transverse  strength,  and  also  the  moduli 
of  elasticity  as  determined  in  the  tensile  tests. 

The  specimens  used  for  these  tensile  tests  were  o".47  in. 
thick,  2.75  in.  wide  in  the  part  between  the  shoulders,  23.6  in. 
long  over  all,  and  about  17  in.  long  between  the  shoulders. 

The  specimens  used  for  compression  were  3.54  in.  long  and 
1. 1 8  in.  square  in  section. 

The  specimens  for  transverse  strength  were  47.24  in.  long, 
5.5  in.  deep,  and  2.17  in.  thick,  the  span  used  being  39.37  in., 
and  the  load  being  applied  at  the  middle. 


458 


APPLIED  MECHANICS. 


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TENSILE  AND  COMPRESSIVE  STRENGTH  OF  STEEL.    459 

The  second  set  of  tests  quoted  will  be  some  made  on  the 
government  testing-machine  at  Watertown  Arsenal,  upon  speci- 
mens of  steel  containing  different  percentages  of  carbon,  the 
tests  themselves  forming  a  portion  of  a  series  denominated  in 
the  government  reports  as  the  "  Temperature  Series."  The 
account  of  the  tests  to  be  quoted  is  to  be  found  in  their  report 
for  1887,  while  the  following  description  of  the  steel  is  taken 
from  that  for  1886;  and  includes  also  some  conclusions  from 
some  earlier  tests  made  there  in  1886  :  — 

"  Ten  grades  of  open-hearth  steel  are  here  represented,  in 
which  the  carbon  ranges  from  0.09  to  0.97  per  cent,  varying  by 
tenths  of  a  per  cent  as  nearly  as  was  practicable  to  obtain  the 
steel. 

"  The  other  elements  do  not  follow  any  regular  succession. 

"  The  bars  were  rolled  to  i\  in.  diameter  and  turned  to 
specimens  from  that  size. 

"  The  steel  was  made  by  the  Norway  Iron  Works,  South 
Boston,  Mass.,  and  the  chemical  analyses  furnished  by  George 
H.  Billings,  superintendent. 

"  These  bars  are  intended  to  furnish  part  of  the  material 
for  use  in  a  comprehensive  series  of  tests  at  higher  tempera- 
tures. 

"  The  modulus  of  elasticity  in  compression  averages  higher 
than  in  tension,  and  appears  less  regular  under  the  former 
stress  than  the  latter. 

"  In  the  case  of  tension  bars,  which  afforded  the  best  oppor- 
tunities for  accurate  measurements,  it  is  at  least  evident  that 
the  modulus  of  elasticity  is  not  far  from  30000000  pounds 
per  sq.  in.,  whether  the  steel  contains  o.i  per  cent  or  i.oo  per 
cent  of  carbon,  other  elements  at  the  same  time  being 
present." 

The  following  gives  a  description  of  the  mode  of  mak- 
ing the  tests  of  1887,  and  an  explanation  of  the  table  of 
tests:  — 


4^0  APPLIED   MECHANICS. 

"  The  specimens  were  taken  from  bars  \\  in.  diameter  of 
open-hearth  steel  as  described  in  Report  of  1886. 

"  The  total  length  of  the  specimens  was  36"  in.,  with  ends 
ij  in.  diameter,  threaded  IO  per  inch,  Franklin  Institute  stan- 
dard shape.  The  stems  'vere  turned  a  length  of  32  in.,  of 
which  30  in.  was  used  a.  t*ie  gauged  length,  where  the  bars 
were  reduced  to  I.OOQ2  ":.,  diameter.  In  this  condition  the 
modulus  of  elasticity  of  2ich  was  determined  as  recorded  in 
Report  of  1886. 

"  After  a  period  of  rest  the  tests  were  resumed,  and  carried 
to  final  rupture,  the  stems  now,  with  the  exception  of  Nos.  i, 
2,  and  the  first  bar  of  No.  7,  being  turned  down  to  smaller 
diameters. 

"  The  first  bar,  No.  7,  broke  in  the  head  at  the  root  of  the 
thread,  and  another  bar  was  prepared  with  the  diameter  of  the 
stem  reduced  to  0.757  in. 

"  Less  ductile  metals  require  a  greater  difference  between 
the  areas  of  the  stem  and  at  the  root  of  the  thread  in  the 
head,  to  avoid  rupture  of  the  latter,  tlian  is  required  by  metal 
possessing  greater  ductility. 

"  It  is  also  shown  in  bars  Nos.  8,  9,  and  10  how  the  centre- 
punch  marks,  which  were  about  o.oi  in.  deep,  in  the  surface  of 
the  stems  located  the  places  of  rupture. 

"  With  increased  hardness  of  metal  slight  imperfections 
become  of  serious  importance. 

"  A  diagram  is  given  (in  the  report)  with  curves  repre- 
senting the  behavior  of  the  several  grades  of  steel  from 
the  initial  loads  up  to  the  time  the  tensile  strength  was 
reached. 

"  When  loads  are  slowly  applied,  the  rupture  of  duc- 
tile metal  does  not  occur  under  the  maximum  stresses,  but 
while  local  contraction  of  area  is  going  on,  and  the  load  is 
falling. 


TENSILE  AND  COMPRESSIVE  STRENGTH  OF  STEEL.    461 

"  The  contraction  of  area  usually  progresses  at  a  more 
rapid  rate,  however,  than  the  load's  fall,  which  results  in  a 
higher  strength  per  square  inch,  measured  on  the  ruptured 
section,  than  the  tensile  strength,  which  is  referred  to  the 
original  diameter  of  the  bar. 

"  Brittle  metal  breaks  at  once,  with  little  or  no  warning 
under  the  maximum  load. 

"  The  mechanical  work  done  is  computed,  beginning  with 
the  initial  load,  and  values  are  given,  in  the  tabulation,  which 
indicate  the  amount  of  work  done  at  the  elastic  limit,  also 
when  the  maximum  load  or  tensile  strength  was  reached. 

"  For  those  bars  which  failed  suddenly  and  the  elongation 
was  not  determined  at  the  maximum  load,  the  next  lower 
load  and  elongation  is  used  in  computing  the  mechanical 
work. 

"  It  is  seen  that  bars  having  the  highest  tensile  strength 
were  not  those  which  required  the  most  work  to  cause  rup- 
ture. 

"  A  number  of  bars  show  a  yielding-point  immediately  upon 
passing  the  elastic  limit,  during  which  period  the  metal  stretches 
and  takes  a  considerable  set  under  loads  below  those  which  had 
previously  been  sustained. 

"  In  an  earlier  report  results  were  shown  in  which  this 
behavior  was  repeated  in  the  same  bars  as  higher  loads  were 
reached." 

The  table  of  results  is  as  follows  :  — 


462 


APPLIED    MECHANICS. 


TENSILE   TESTS   OF   STEEL    BARS— TEMPERATURE    SERIES. 

Tests  at  Atmosplieric  Temperature. 


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The  third  set  of  tests  that  will  be  quoted  in  this  connection 
is  one  made  by  the  Committee  on  Chemical  Research  of  the 
United  States  Government  Commission,  and  recorded  in  Ex- 
ecutive Document  23,  46th  Congress,  second  session. 

The  following  summary  of  this  report  will  be  appended 
here,  giving  only  the  percentages  of  the  most  important  ingre- 
dients besides  the  iron.  This  summary  is  as  follows : 


TENSILE  STRENGTH  OF  STEEL. 


463 


Laboratory  Number. 
3  Samples  of  Each. 

Average  Elastic  Limit, 
in  Ibs.,  per  Sq.  In. 

Average  Modulus  of 
Elasticity. 

Average  Breaking- 
Strength  per  Sq.  In. 
of  Original  Area. 

Average  Breaking- 
Strength  per  Sq.  In. 
of  Fractured  Area. 

Chemical  Analysis. 

Combined 
Carbon. 

Graphitic 
Carbon. 

j 

I 
cn 

Phosphorus. 

| 

1518 

21600 

24648000 

37933 

96033 

0.246 

O.OII 

0.145 

0.004 

0.014 

% 
0.02O 

1519 

40000 

28010000 

67666 

109166 

I  0.383 
1  0.389 

>  0.006 

0.076 

0.003 

0.017 

O.o6o 

1520 

37052 

26578000 

56659 

123433 

(  0.276 
\  o-277 

|  0.018 

0.061 

O.OOI 

0.015 

O.O3I 

1521 

44577 

26750000 

67817 

139633 

\  o-377 
f  0.372 

j  0.025 

0.070 

O.002 

0.014 

O.O27 

1522 

35854 

25799000 

60214 

121800 

(  0.279 
I  o-273 

|  0.022 

0.070 

0.002 

0.015 

O.O4I 

1523 

41721 

26507000 

78119 

114300 

(  0.468 
|  0.464 

j  0.027 

0.090 

0,001 

0.015 

0.032 

1524 

45090 

26013000 

83497 

129933 

j  0.728 
t  0.724 

/  O.O28 

0.126 

0.002 

0.014 

0.048 

1525 

49544 

24816000 

90940 

122133 

(  0.611 
{  0.608 

j  0.027 

O.OIO 

O.003 

0.014 

0.042 

1526 

51282 

24588000 

95340 

130600 

(  0.676 
\  0.678 

|  O.OlS 

0.135 

0.003 

0.015 

0.038 

1527 

51609 

27049060 

100719 

121900 

\  0-557 
(  0-551 

\  0.027 

O.II2 

O.OO2 

0.015 

0.028 

1528 

51076 

26587000 

115169 

I38I33 

j  0.833 
(  0.828 

>  0.040 

0.134 

Trace 

0.015 

0.046 

1529 

523H 

25979000 

120275 

I4I433 

(  0.908 
}  0.909 

j  0.015 

O.I4I 

0.002 

0.014 

0.036 

1530 

48597 

26648000 

123697 

144492 

0.924 

0.022 

0.187 

0.002 

0.014 

0.031 

1531 

55954 

24682000 

118156 

140367 

0.966 

0.030 

0.153 

Trace 

0.015 

0.035 

1532 

54596 

26819000 

117660 

141867 

0.946 

0.030 

O.l62 

O.OOI 

0.015 

0.029 

1533 

60627 

26250000 

116636 

138600 

1.024 

0.020 

0.173 

Trace 

0.013 

0.014 

1534 

59160 

25576000 

110823 

118500 

1.079 

0.027 

0.096 

O.OOI 

0.014 

0.044 

1535 

66710 

27091000 

120602 

132060 

1.  112 

0.030 

O.igO 

Trace 

0.015 

0.045 

1536 

63885 

25186000 

"95M 

128800 

1.186 

0.024 

O.II4 

Trace 

0.014 

0.132 

1537 

70405 

25824000 

121253 

128800 

1.285 

0.033 

0.106 

Trace 

0.015 

0.273 

1053 

58971 

28169000 

103761 

149400 

f  0.973  -j 

0.213 

0.003 

0.025 

0.073 

1054 

59040 

28608000 

106205 

167367 

0.886 

0.196 

0.002 

0.037 

0.185 

1055 

51500 

31368000 

92081 

151467 

g    0.694 

0.128 

Trace 

0.037 

0.137 

1056 

57802 

27668000 

94529 

137900 

•§    0.994 

0.140 

0.003 

0.027 

O.IOI 

1057 

50494 

27776000 

72979 

I5H33 

(J  <   0.401 

0.085 

0.006 

0.032 

O.II2 

1058 

54763 

26574000 

96373 

152233 

3   0.905 

0.161 

Trace 

0.026 

0.108 

io59 

54243 

29028000 

95612 

127133 

H    0.915 

0.191 

0.002 

0.026 

0.086 

1060 

50286 

25538000 

70066 

102967 

0.238 

0.105 

0.012 

0.034 

0.184 

1061 

40667 

24536000 

67000 

120067 

0.463  ' 

O.I2I 

0.002 

O.O2O 

Trace 

1065 

35167 

26379000 

55000 

140600 

0.184 

0.009 

0.063 

Trace 

0.014 

0.051 

1066 

48367 

25637000 

83660 

139267 

0-459 

O.IlS 

0.108 

Trace 

O.O26 

0.185 

1067 

,50583 

28546000 

81216 

150367 

0-451 

O.OO3 

0.134 

0.004 

O.O26 

0.139 

464 


APPLIED   MECHANICS. 


Laboratory  Number. 
3  Samples  of  Each. 

Average  Elastic  Limit, 
in  lbs.,per  Sq.  In. 

Average  Modulus  of 
Elasticity. 

Average  Breaking- 
Strength  per  Sq.  In. 
of  Original  Area. 

Average  Breaking- 
Strength  per  Sq.  In. 
of  Fractured  Area. 

Chemical  Analysis. 

Combined 
Carbon. 

Graphitic 
Carbon. 

Cfl 

I 
in 

Phosphorus. 

Manganese. 

1068 

50522 

28755000 

111910 

137233 

0-793 

0.013 

0.172 

None 

% 
0.019 

0.193 

1069 

5*435 

25627000 

117668 

143633 

0.825 

0.009 

0.170 

None 

0.020 

0.182 

1070 

55*37 

26498000 

126577 

*5*433 

0.846 

0.007 

0.233 

None 

O.Oig 

0.191 

1071 

62691 

28219000 

129837 

151800 

0-959 

0.013 

0.225 

Trace 

O.OlS 

0.213 

1072 

61062 

27643000 

121416 

148900 

0984 

0.008 

0.157 

None 

O.Oig 

0.245 

1073 

68015 

26805000 

121920 

132000 

1.059 

0.013 

0.162 

None 

0.022 

0.252 

1074 

69427 

27552000 

126036 

144300 

1.108 

0.013 

0.206 

None 

0.023 

0.269 

1075 

72700 

27760000 

129293 

145567 

1.142 

0.012 

0.204 

None 

O.O2O 

0.282 

1076 

75403 

27264000 

132010 

138600 

1.244 

0.082 

0.246 

None 

O.OI7 

0.262 

1077 

50000 

28102000 

96699 

101800 

0.627 

0.012 

0.154 

Trace 

O.OO7 

0.050 

1078 

49436 

28701000 

9*538 

i69733 

0-439 

O.OI5 

0.136 

O.OOI 

O.OI9 

0.023 

1079 

3694* 

27525000 

80945 

135*67 

0.461 

None 

0.116 

Trace 

O.O2O 

0.027 

.  1080 

25167 

25953000 

434*7 

88500 

(  0.209  =  total  ) 
|   carbon.   \ 

0.163 

0.009 

0.084 

O.O2O 

1081 

28333 

29028000 

5*833 

149200 

0.116 

0.014 

O.OII 

0.029 

0.045 

0.192 

1082 

5*609 

28518000 

102240 

109333 

0.675 

0.016 

0.028 

0.028 

0.065 

0-459 

1083 

61388 

26295000 

128750 

139833 

0.670 

O.OII 

0.043 

t>.oi4 

0.063 

0.625 

1084 

34333 

24589000 

54333 

99800 

0.049 

0.008 

0.219 

0.007 

0.179 

0.063 

1085 

48350 

28752000 

85833 

155267 

0-433 

0.014 

o  071 

0.008 

O.o62 

0.493 

1086 

44667 

25360000 

65000 

13*833 

O.I  II 

O.OI2 

0.006 

0.031 

0.123 

o  103 

1087 

542*7 

25400000 

88007 

92967 

0.348 

0.007 

0.028 

0.041 

0.125 

0.404 

1088 

65190 

23648000 

101045 

102300 

0.744 

O.OI2 

0.074 

0.043 

0.104 

0.465 

1089 

27617 

23341000 

48082 

97033 

0.042 

None 

o.i  68 

O.OII 

0.109 

0.051 

1090 

30000 

24071000 

5*5oo 

75833 

0.012 

0.003 

0.214 

0.048 

0-3*5 

0.081 

1091 

26000 

25482000 

59000 

80033 

0.230 

0.004 

0.084 

Trace 

0.039 

None 

1092 

54869 

25503000 

793*5 

84733 

0-3*4 

0.012 

0.016 

0.066 

0.099 

0-525 

1093 

48733 

29326000 

78500 

92167 

0.243 

O.OII 

0.013 

0.058 

0.128 

o-34* 

1094 

47833 

26110000 

68667 

88467 

0.135 

0.004 

0.007 

0.056 

O.II3 

0.165 

1095 

56505 

25917000 

9534* 

101633 

0-374 

0.008 

0.018 

0.062 

0.138 

0.584 

1096 

63126 

24602000 

111203 

142133 

0-375 

0.012 

0.070 

0.038 

0.092 

0.685  ' 

1582 

379*9 

28697000 

55955 

182533 

- 

- 

- 

- 

- 

- 

1583 

33247 

28205000 

50523 

161700 

:-V 

- 

~ 

- 

- 

1588 

60627 

27010000 

110389 

162900 

J-V 

'  ~ 

- 

- 

- 

- 

1589 

58562 

25790000 

106695 

176467 

-•*v. 

- 

- 

- 

~ 

- 

1590 

50305 

25772000 

79858 

154267 

':"'•' 

;"A*'  : 

- 

- 

~ 

- 

1591 

60627 

27389000 

123645 

142533 

- 

•i-" 

- 

- 

- 

- 

1592 

58345 

25323000 

103761 

147167 

- 

- 

- 

- 

- 

- 

1593 

30933 

24633000 

49167 

956oo 

- 

- 

- 

ft  -^\.: 

- 

- 

1594 

53565 

25776000 

78608 

148300 

- 

- 

- 

'"--*'  ' 

- 

- 

1595 

46394 

25330000 

64104 

129200 

" 

" 

" 

' 

TENSILE   STRENGTH  OF  STEEL. 


465 


Mr.  A.  F.  Hill,  in  the  "  Transactions  of  the  Engineers' 
Society  of  Western  Pennsylvania"  for  1880,  gives  an  account 
of  some  tests  of  open-hearth  steel  manufactured  by  Anderson 
&  Co.  of  Pittsburgh,  taken  from  different  runs,  and  ranging 
from  0.3  to  0.5  per  cent  of  carbon  ;  the  steel  being  tested  both 
in  the  form  of  specimens,  and  also  in  the  form  of  eye-bars, 
plates,  and  riveted  plate-girders. 

The  following  account  of  some  of  these  tests  is  condensed 
from  his  paper  :  — 


I.     SPECIMEN   TESTS   OF   BAR-ENDS   30   INCHES    LONG,   DESIGNED    FOR 

EYE-BARS. 


Tensile  Stress  per  Square  Inch 

Mark,  and  Carbon 

at 

Stretch  at 

Reduction 

Percentage. 

Fracture,  %. 

of  Area  at 
Fracture,  %. 

Elastic  Limit. 

Rupture. 

I 

S57I2 

94760 

'S-1 

30 

2 

56009 

95380 

12.9 

26 

3 

0.3%  C. 

55120 

95830 

!5-3 

31 

4 

55830 

96020 

14.5 

27 

5  , 

55512 

94970 

13.8 

29 

i 

65790 

112340 

10.8 

19 

2 

66040 

112470 

8.9 

16 

3 

0.5%  C. 

66160 

111980 

10.5 

22 

4 

65550 

113320 

10.9 

21 

5 

65980 

113040 

94 

20 

466 


APPLIED   MECHANICS, 


II.     EXPERIMENTS    ON    STEEL    EYE-BARS    WITH    0.3%   CARBON. 

Dimensions  :  Stem,  3  in.  X  f  in.  X  10  ft.     Head,  ij  in.  thick,  7^  in.  across  the  eye. 
Pin-hole,  3!  in.  diameter. 


1 

Tensile  Stress  per 

I 

I* 

Square  Inch  at 

£ 

•s  £ 

•w 

3 

• 

Mark. 

*O     4> 

^ 

O     ° 

Remarks. 

ll 

Elastic 
Limit. 

Rupture. 

1  a 

j;    3 

1    ^ 
•g     « 

* 

^ 

K 

Bl 

54026 

87400 

8.2 

44 

Broke  i  ft.  3    in.  from  pin-hole. 

B2 

a 

54H3 

94500 

9.2 

46 

"      4  ft.  3!  in.      " 

B3 

13 

54H3 

89300 

7.0 

42 

"      i  ft.  8i  in.     « 

Fl 

T3 

51762 

92672 

8.2 

o 

Broke  in  head  into  3  pieces. 

F2 

'o 

54065 

9J570 

9-3 

29 

"      5  ft.  2\  in.  from  pin-hole. 

F3 

52518 

94780 

1  1.8 

40 

"      2  ft.  if  in.     "          " 

Al 

•d 

^<u 

58473 

69140 

2.0 

i>    "o 

3     c     G 

Broke  5^  in.  from  pin-hole. 

A2 

13 

56059 

63000 

2.6 

tn    £    v 

"      5^  in.     "          " 

A3 

^ 

55310 

69400 

2.2 

si2 

"      5f  in.     " 

In  those  marked  "upset,"  the  bars  were  rolled  to  the  re- 
quired section  of  the  stem,  with  sufficient  surplus  of  length 
to  form  the  heads  by  hydraulic  upsetting. 

In  those  marked  "rolled,"  the  heads  were  rolled  by  Klo- 
man's  patent  process. 

In  those  marked  "  welded,"  the  heads  were  formed  by  weld- 
ing pieces,  and  die  forging. 

Fractures  of  Bi,  B2,  63,  F2,  and  F3  were  fine,  silky,  and 
wedge-shaped. 

Fi  broke  in  head,  and  showed  effect  of  over-heating;  frac- 
ture coarse  and  granular. 

A  i,  A2,  and  A3  broke  in  stem  close  to  neck  ;  fracture  close- 
grained,  50  per  cent  granular,  and  showing  effect  of  welding 


TENSILE  STRENGTH  OF  STEEL.  467 

heat.  Weld  was  defective  at  junction  of  head  and  neck  on 
account  of  welding-pieces  having  been  too  small. 

Mr.  Hill  draws  from  these  tests  the  following  conclusions  :  — 

i°.  "The  strength  of  the  specimen  exceeds,  in  each  case, 
that  of  the  manufactured  bar." 

2°.  "  The  uniformity  of  the  results  obtained  from  the  tests 
of  the  bar-ends  shows  conclusively  that  whatever  difference  in 
strength  there  is  between  these  bar-ends  and  the  manufactured 
eye-bar  is  properly  ascribable  to  the  mode  of  manufacture." 

3°.  "The  results  obtained  from  the  rolled  and  the  upset  eye- 
bars  approach  nearest  to  the  original  bar  strength,  and  give  the 
best  results.  The  difference  between  the  results  from  these  two 
methods  is  so  trifling  —  and  if  any  thing  in  the  0.3  per  cent  car- 
bon group,  slightly  in  favor  of  the  upset  bar  —  that  it  leaves  no 
doubt  in  my  mind  that  these  two  processes  are  equally  good." 

4°.  "  The  results  from  the  welded  bars  show,  that,  while 
steel  can  be  perfectly  welded,  there  is  a  loss  of  nearly  30  per 
cent  of  ultimate  strength  as  compared  with  the  original  bar ; 
moreover,  the  elastic  limit  is  too  near  the  ultimate  strength,  and 
the  percentage  of  elongation  too  small,  to  give  sufficient  warn- 
ing of  impending  failure." 

"  It  will,  therefore,  be  safe  to  conclude  that  welded  members 
in  steel  construction,  while  no  worse  than  welded  iron  ones,  are 
not  desirable,  and,  in  fact,  ought  not  to  be  admitted  at  all, 
except  where  the  grade  of  steel  used  is  very  low ;  and  then  the 
greatest  caution  in  working  and  annealing  will  be  required." 

Mr.  Hill  states  also  that  his  experiments  with  eye-bars  con- 
taining 0.5  per  cent  carbon  bear  out  these  conclusions. 

Tests  of  Steel  Eye-Bars  made  on  the  Government  Machine.  — 
In  Executive  Document  No.  5,  48th  Congress,  first  session,  is 
the  record  of  the  tests  of  six  eye-bars  of  steel,  presented  by 
the  president  of  the  Keystone  Bridge  Company. 

The  following  is  an  extract  from  the  report  in  regard  to 
these  eye-bars  :  — 


468 


APPLIED  MECHANICS. 


"  The  eye-bars  were  made  of  Pernot  open-hearth  steel,  fur- 
nished by  the  Cambria  Iron  Company  of  Johnstown,  Penn. 

"The  furnace  charges,  about  15  tons  each  of  cast-iron, 
magnetic  ore,  spiegeleisen,  and  rail-ends,  preheated  in  an  aux- 
iliary furnace,  required  six  an^d  one-half  hours  for  conversion. 

"  All  these  bars  were  rolled  from  the  same  ingot. 

"  Samples  were  tested  at  the  steel-works  taken  from  a  test 
ingot  about  one  inch  square,  from  which  were  rolled  |-inch 
round  specimens. 

"The  annealed  specimen  was  buried  in  hot  ashes  while  still 
red-hot,  and  allowed  to  cool  with  them. 

"The  following  results  were  obtained  by  tensile  tests  :  — 


Elastic 
Limit,  in 
Ibs.',  per 
Sq.  In. 

Ultimate 
Strength,  in 
Ibs.,  per 
Sq.  In. 

Contrac- 
tion of 
Area. 

Modulus 
of 
Elasticity. 

Carbon. 

f-inch  round  rolled  bar  . 

48040 

73150 

%• 

45-7 

28210000 

%- 
0.27 

f-inch  round  rolled  and 

annealed  bar  .... 

42210 

69470 

54-2 

29210000 

0.27 

"The  billets  measured  7  inches  by  8  inches,  and  were 
bloomed  down  from  14-inch  square  ingot. 

"They  were  rolled  down  to  bar-section  in  grooved  rolls  at 
the  Union  Iron  Mills,  Pittsburgh. 

"The  reduction  in  the  roughing-rolls  was  from  7  inches  by 
8  inches  to  6\  inches  by  4  inches ;  and  in  the  finishing-rolls,  to 
6^  inches  by  I  inch. 

"  The  eye-bar  heads  were  made  by  the  Keystone  Bridge 
Company,  Pittsburgh,  by  upsetting  and  hammering,  proceeding 
as  follows  :  — 

"  The  bar  is  heated  bright  red  for  a  length  of  (approxi- 
mately) 27  inches,  and  upset  in  a  hydraulic  machine ;  after 


TENSILE  STRENGTH  OF  STEEL.  469 

which  the  bar  is  reheated,  and  drawn  down  to  the  required 
thickness,  and  given  its  proper  form  in  a  hammer-die. 

"  The  bars  are  next  annealed,  which  is  done  in  a  gas-furnace 
longer  than  the  bars.  They  are  placed  on  edge  on  a  car  in  the 
annealing-furnace,  separated  one  from  another  to  allow  free 
circulation  of  the  heated  gases.  They  are  heated  to  a  red  heat, 
when  the  fires  are  drawn,  and  the  furnace  allowed  to  cool. 
Three  or  four  days,  according  to  conditions,  are  required  before 
the  bars  are  withdrawn. 

"  The  pin-holes  are  then  bored. 

"  The  analyses  of  the  heads  before  annealing  were  :  — 

"  Carbon,  by  color 0.270  per  cent. 

Silicon 0.036        " 

Sulphur o-°75        " 

Phosphorus 0.090       " 

Manganese 0.380       " 

Copper Trace. 

"  The  bars  were  tested  in  a  horizontal  position,  secured  at 
the  ends,  which  were  vertical. 

"  To  prevent  sagging  of  the  stem,  a  counterweight  was  used 
at  the  middle  of  the  bar. 

"  Before  placing  in  the  testing-machine,  the  stem  from  neck 
to  neck  was  laid  off  into  lo-inch  sections,  to  determine  the 
uniformity  of  the  stretch  after  the  bar  had  been  fractured. 

"A  number  of  intermediate  ic-inch  sections  were  used  as 
the  gauged  length,  obtaining  micrometer  measurements  of 
elongation,  and  the  elastic  limit  for  that  part  of  the  stem  which 
was  not  acted  upon  during  the  formation  of  the  heads.  Elon- 
gations were  also  measured  from  centre  to  centre  of  pins,  taken 
with  an  ordinary  graduated  steel  scale. 

"  The  moduli  of  elasticity  were  computed  from  elongations 
taken  between  loads  of  10000  and  30000  Ibs.  per  square  inch, 
deducting  the  permanent  sets. 


47°  APPLIED   MECHANICS. 

"The  behavior  of  bars  Nos.  4582  and  4583,  after  having 
been  strained  beyond  the  elastic  limit,  is  shown  by  elongations 
of  the  gauged  length  measured  after  loads  of  40000  and  50000 
Ibs.  per  square  inch  had  been  applied ;  and  with  bar  No.  4583, 
after  its  first  fracture  under  64000  Ibs.  per  square  inch,  a  rest 
of  five  days  intervening  between  the  time  of  fracture  and  the 
time  of  measuring  the  elongations. 

"Considering  the  behavior  between  loads  of  10000  and 
30000  Ibs.  per -square  inch,  we  observe  the  elongations  for  the 
primitive  readings  are  nearly  in  exact  proportion  to  the  incre- 
ments of  load. 

"  Loads  were  increased  to  40000  Ibs.  per  square  inch,  passing 
the  elastic  limit  at  about  37000  Ibs.  per  square  inch ;  the  respec- 
tive permanent  stretch  of  the  bars  being  1.31  and  1.26  per  cent. 

"Elongations  were  then  immediately  redetermined,  which 
show  a  reduction  in  the  modulus  of  elasticity,  as  we  advanced 
with  each  increment,  of  5000  Ibs.  per  square  inch. 

"  Corresponding  measurements  after  the  bars  had  been 
loaded  with  50000  Ibs.  per  square  inch  reach  the  same  kind  of 
results. 

"The  first  fracture  of  bar  No.  4583,  under  64000  Ibs.  per 
square  inch,  occurred  at  the  neck,  leaving  sufficient  length  to 
grasp  in  the  hydraulic  jaws  of  the  testing-machine,  and  con- 
tinue observations  on  the  original  gauged  length.  This  was 
done  after  the  fractured  bar  had  rested  five  days. 

"The  elongations  now  show  the  modulus  of  elasticity  con- 
stant or  nearly  so,  the  only  difference  in  measurements  being 
in  the  last  figures,  up  to  50000  Ibs.  The  readings  were  then 
immediately  repeated,  and  the  same  uniformity  of  elongations 
obtained. 

"  An  illustration  of  the  serious  influence  of  defective  metal 
in  the  heads  is  found  in  the  first  fracture  of  bar  No.  4583. 

"  There  was  about  27  per  cent  excess  of  metal  along  the 
line  of  fracture  over  the  section  of  the  stem." 


TENSILE  STRENGTH  OF  STEEL. 

* 


471 


Ho    I5 

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Gauged  Length, 

in  inches. 

1  S  C  C    •     • 

ON        ON 

Width,  in  inches. 

O        0        0        0 

P         P 

Hi 

3    S3    %   %       '        ' 

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Maximum      Compression 
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472 


APPLIED   MECHANICS. 


ELONGATIONS    OF    No.  4582    FOR  EACH    INCREMENT   OF   5000   LBS.   PER 

SQUARE    INCH. 


Loads,  in  Ibs., 
per 
Square  Inch. 

Elongations. 

Primitive  Load- 
ing. 

After  Load  of 
40000  Ibs.  per 
Square  Inch. 

After  Load  of 
50000  Ibs.  per 
Square  Inch. 

IOOOO 

. 

I5OOO 
20000 
25000 
30000 

0.0274 
0.0269 
0.0269 
0.0269 

0.0300 
0.0305 
0.0320 
0.0330 

0.03II 
O.O322 
0-0337 
0.0341 

ELONGATIONS    OF   No.  4583    FOR  EACH    INCREMENT   OF   5000  LBS.   PER 

SQUARE    INCH. 


Loads,  in 

Elongations. 

Elongations  after  64000  Ibs.  per 
Square  Inch. 

Ibs.,  per 

Square  Inch. 

Primitive  Load- 

After 40000  Ibs. 

After  50000  Ibs. 

First 

Second 

ing. 

per 
Square  Inch. 

per 
Square  Inch. 

Reading. 

Reading. 

IOOOO 

. 

. 

_ 

_ 

15000 

0.0272 

0.0291 

0.0302 

0.0311 

0.0310 

20000 

0.0272 

0.0305 

0.0315 

0.0308 

0.0310 

•  25000 

0.0268 

0-0,314 

0.0325 

0.03II 

0.0310 

30000 

0.0267 

0.0326 

0.0340 

0.0312 

0.0310 

35000 

- 

- 

- 

0.0311 

- 

4OOOO 

- 

- 

- 

0.0312 

- 

45000 

- 

- 

- 

0.0310 

- 

5OOOO 

; 

" 

' 

0.0315 

STEEL  PLATES. 


Steel   plates  are  used  in   making  plate-girders  and 
forms  for  resisting  load,  and  also  for  steam-boilers. 


other 


STEEL  PLATES. 


473 


For  the  latter  purpose  the  steel  must  be  very  low  in  carbon, 
and  must  stand  very  "much  more  in  the  way  of  bending,  punch- 
ing, shearing,  etc.,  than  in  the  former  case.  Its  tensile  strength 
will  not  be  high,  but  it  must  have  great  ductility.  The  first  set 
of  tests  to  which  reference  will  be  made  is  the  set  given  in  Mr. 
Hill's  paper  already  referred  to,  this  steel  being  too  high  to  be 
suitable  for  boiler-work. 

He  had  made  54  tests  of  rolled  open-hearth  steel  plates. 

The  following  15  tests  give  the  relative  strengths  of  the 
plates,  and  the  percentages  of  carbon. 

The  plates  tested  were  all  f  inch  thick,  12  inches  wide,  and 
6  feet  long,  tested  as  they  came  from  the  rolls  ;  crop  ends 
sheared,  50  inches  between  jaws  of  machine. 


Tensile  Stress,  in  Ibs.,  per 

Square   Inch,    in   Direction 

Average,  per 

Mark. 

Carbon, 

of  Rolling,  at 

cent, 

Remarks. 

percentage. 

Elongation  at 

Fracture. 

Elastic  Limit. 

Rupture. 

Pi 

o-3 

43260 

79120 

?2 

o-3 

44820 

77840 

P3 

0-3 

45IIO 

78390 

'      19-3 

Fractures  fine  and  silky. 

P4 

o-3 

43990 

77970 

P5 

°-3 

44720 

78280 

^ 

Ri 

0.4 

51620 

81990 

R2 

0.4 

50980 

81720 

R3 

0.4 

51260 

8373Q 

'     13-9 

Fractures  very  fine. 

R4 

0.4 

5IIOO 

81830 

RS 

0.4 

50890 

8313° 

Vi 

0-5 

58950 

85790 

V2 

V3 
V4 

o-5 
o-5 
o-5 

59200 
58540 
58880 

86220 
85560 
86000 

10.5 

(Fractures  good,  slightly 
I     granular  on  edges. 

vs 

°-5 

59330 

86330 

474 


APPLIED   MECHANICS. 


He  then  gives  the  following  table : 


COMPARATIVE    RESULTS    OF   SHEARING,  PUNCHING,  ANNEALING,  AND 
TEMPERING    STEEL    PLATE. 


Carbon, 
per  cent. 


Treatment  of  Specimen. 


Cut  in  planer 49431 

0.3        Sheared 3237o 

Punched  .     .* o 

Punched  and  hammered  cold      .     .  o 

Punched,  hammered,  and  annealed,  55780 

0.4        Cut  in  planer 63475 

0.4        Sheared 46900 

0.4        Punched o 

0.4        Sheared  and  annealed 5935° 

0.4        Punched  and  tempered 52780 

Cut  in  planer 65186 

Sheared 51666 

Punched  o 

Sheared  and  tempered 60375 

Punched  and  annealed 57960 


Tensile  Stress  per  Square 
Inch  at 


Elastic  Limit.       Rupture. 


94396 
74980 
63410 
87540 
100410 
87695 

7533° 
68890 
86160 
103560 
84092 
79000 
78400 
87293 
84900 


From  these  tests  he  draws  the  following  conclusions  :  — 

1°.  "That  both  shearing  and  punching  are  injurious  to  all 
grades  of  steel,  and  cold  punching  far  more  than  shearing." 

2°.  "  That  both  these  operations  affect  the  elastic  limit  far 
more  than  they  do  the  ultimate  strength." 

3°.  "That  apparently  the  lower  grades  of  steel  are  more  in- 
jured than  higher  grades ;  but  the  evidence  on  this  point  is  not 
certain,  as  the  lower-grade  plates  were  thicker  than  those  of 
higher  grade." 


COLD    CRYSTALLIZATION  OF  IRON  .AND    STEEL.        475 


COLD   CRYSTALLIZATION   OF  IRON  AND   STEEL. 

The  question  of  cold  crystallization  of  wrought-iron  and 
steel  is  one  that  has  been  agitated  from  the  earliest  times,  and, 
although  Kirkaldy  tried  to  dispose  of  it  finally  by  offering  evi- 
dence showing  that  it  does  not  exist,  nevertheless  we  find  the 
same  old  question  cropping  out  every  little  while,  and  although 
the  bulk  of  the  evidence  is  admitted  to  be  against  it,  and,  as  it 
seems  to  the  writer,  there  is  no  evidence  in  its  favor,  we  find 
every  now  and  then  some  one  who  thinks  that  certain  observed 
phenomena  can  be  explained  in  no  other  way. 

The  most  usual  phenomenon  which  cold  crystallization  is 
called  upon  to  explain  is  the  crystalline  appearance  of  the 
fracture  of  some  piece  of  wrought-iron  or  steel  that  has  been 
in  service  for  a  long  time,  and  which  has,  as  a  rule,  been  sub- 
jected to  more  or  less  jars  or  shocks.  The  cases  most  fre- 
quently cited  are  those  of  axles  of  some  sort  which  have  been 
broken,  and,  in  the  case  of  which,  the  fracture  has  had  a  crys- 
talline appearance,  and  where  samples  cut  from  the  other  parts 
of  the  axle  and  tested  have  shown  a  fibrous  fracture. %  The 
assumption  has  therefore  been  made  that  the  iron  was  origi- 
nally fibrous,  and  that  crystallization  has  been  caused  by  the 
shocks  or  the  jarring  to  which  it  has  been  subjected  in  the 
natural  service  for  which  it  was  intended. 

Kirkaldy  showed  (see  his  sixty-six  conclusions)  that  when 
fibrous  iron  was  broken  suddenly,  or  when  the  form  of  the 
piece  was  such  as  not  to  offer  any  opportunity  for  the  fibres  to 
stretch,  the  fibres  always  broke  off  short,  and  the  fracture  was 
at  right  angles  to  their  length,  and  hence  followed  the  crystal- 
line appearance ;  whereas  if  the  breaking  was  gradual,  and  the 
fibres  had  a  chance  to  stretch,  they  produced  a  fibrous  appear- 
ance :  in  short,  he  claimed  that  the  difference  between  the  crys- 
talline or  the  fibrous  appearance  of  the  fracture  was  only  a 


476  APPLIED   MECHANICS. 


difference  of  appearance,  and  not  a  change  of  internal  structure 
from  fibrous  to  crystalline. 

The  facts  that  Kirkaldy  showed  in  tfiis  regard  are  generally 
acknowledged  to-day,  and  doubtless  answer  by  far  the  greater 
part  of  those  who  claimed  cold  crystallization  at  the  time  that 
he  wrote,  and  also  a  great  many  of  those  who  claim  its  exist- 
ence to-day. 

But  it  is  easy,  if  suitable  means  be  taken,  to  distinguish 
cases  of  crystalline  appearance  of  fracture  from  cases  where 
there  are  actual  crystals  in  the  piece  ;  and  it  is  rather  about 
those  cases  where  the  iron  near  the  fracture  actually  contains 
distinct  crystals  that  what  discussion  there  is  to-day  that  is 
worth  considering  takes  place. 

The  number  of  such  cases  is,  of  course,  small,  but  every 
once  in  a  while  some  one  is  cited,  and  the  claim  is  put  forward 
that  the  iron  was  originally  fibrous,  and  that  these  crystals 
must  therefore  have  been  produced  without  heating  the  iron 
to  a  temperature  where  chemical  change  is  known  to  occur. 

Inasmuch  as  the  one  who  claims  the  existence  of  cold  crys- 
tallization is  announcing  a  theory  which  is  manifestly  opposed 
to  the^  well-known  chemical  law  that  crystallization  requires 
freedom  of  molecular  motion,  and  hence  can  only  take  place 
from  solution,  fusion,  or  sublimation,  it  follows  that  the  burden 
of  proof  rests  with  him,  and  before  he  can  substantiate  his 
theory  in  any  single  case  he  must  prove  beyond  the  possibility 
of  doubt,  i°,  that  the  iron  or  steel  was  originally  fibrous,  i.e., 
not  only  that  fibrous  iron  was  used  in  manufacturing  the  pieces, 
but  also  that  it  had  not  been  overheated  during  its  manufac- 
ture, and,  2°,  that  it  has  never  been  overheated  during  its  period 
of  service. 

'  It  is  because  the  writer  is  not  aware  of  any  case  where  these 
two  circumstances  have  been  proved  to  hold  that  he  says  that 
he  knows  of  no  evidence  for  cold  crystallization.  In  this  con- 
nection it  is  not  worth  while  to  quote  very  much  of  the  exten- 


COLD    CRYSTALLIZATION  OF  IRON  AND    STEEL.        477 

sive  literature  on  the  subject ;  hence  only  a  little  of  the  most 
modern  evidence  will  be  given  here. 

On  page  1007  et  seq.  of  the  report  of  tests  on  the  govern- 
ment testing-machine  at  Watertown  Arsenal  for  1885  is  given 
an  account  of  a  portion  of  a  series  of  tests  upon  wrought-iron 
railway  axles,  and  the  following  is  quoted  from  that  report :  — 

"  This  series  of  axle  tests,  begun  September,  1883,  is  carried 
on  for  the  purpose  of  determining  whether  a  change  in  struc- 
ture takes  place  in  a  metal  originally  ductile  and  fibrous  to  a 
brittle,  granular,  or  crystalline  state,  resulting  from  exposure 
to  such  conditions  as  are  met  with  in  the  ordinary  service  of  a 
railway  axle. 

"  Twelve  axles  were  forged  from  one  lot  of  double-rolled 
muck-bars,  and  in  their  manufacture  were  practically  treated 
alike.  Each  axle  was  made  from  a  pile  composed  of  nine  bars, 
each  6  in.  wide,  f  in.  thick,  and  3  ft.  3  in.  long,  and  was  finished 
in  four  heats,  two  heats  for  each  end. 

"  The  forging  was  done  by  the  Boston  Forge  Company  in 
their  improved  hammer  dies,  which  finish  the  axle  very  nearly 
to  its  final  dimensions. 

"  Two  axles  were  taken  for  immediate  test,  to  show  the 
quality  of  the  finished  metal  before  it  had  performed  any  rail- 
way service,  and  serve  as  standards  to  compare  with  the 
remaining  ten  axles,  to  be  tested  after  they  had  been  in 
use. 

"  The  axles  are  in  use  in  the  tender-trucks  of  express  loco- 
motives of  the  Boston  and  Albany  Railroad.  Mr.  A.  B.  Under- 
bill, superintendent  of  motive-power,  contributes  the  axles  and 
furnishes  the  record  of  their  mileage." 

The  results  of  some  measurements  of  deflection  are  given 
concerning  one  of  the  axles  in  tender  134,  after  it  had  run 
95000  miles  ;  and  then  follows  :  — 

"  Regarding  the  axle  for  the  time  being  as  cylindrical,  3.96 


4/8  APPLIED   MECHANICS. 

inches  diameter,  the  modulus  of  elasticity  by  computation  will 
be  28541000  pounds. 

"Applying  this  modulus  to  the  deflections  observed  in  rear 
axle  of  the  rear  trucks  of  tender  No.  150,  the  maximum  fibre 
strain  is  found  to  be  9935  pounds  per  square  inch  when  the 
tender  was  partially  loaded,  and  14900  pounds  per  square  inch 
when  fully  loaded. 

"  Taken  together,  the  tensile  and  compressive  stresses, 
which  are  equal,  amount  to  19870  and  29800  pounds  per  square 
inch  respectively,  as  the  range  of  stresses  over  which  the  metal 
works. 

"  This  definition  of  the  limits  of  stresses  must  be  regarded 
as  approximate.  There  are  influences  which  tend  to  increase 
the  maximum  fibre  strain,  such  as  unevenness  of  the  track,  the 
side  thrust  of  the  wheel-flanges  against  the  rails.  On  the  other 
hand,  the  inertia  of  the  axle,  particularly  under  high  rates  of 
speed,  would  exert  a  restraining  influence  on  the  total  deflec- 
tion. 

"  Nine  tensile  specimens  were  taken  from  each  axle  ;  three 
from  each  end,  including  the  section  of  axle  between  the  box 
and  wheel  bearings,  and  three  from  the  middle  of  its  length. 
They  are  marked  M.B.,  with  the  number  of  the  axle  ;  also  a 
sub-number  and  letter  to  indicate  from  what  part  of  the  axle 
each  was  taken. 

"  The  tensile  test-pieces  showed  fibrous  metal,  and  generally 
free  from  granulation. 

"  The  muck-bar  had  a  higher  elastic  limit  and  lower  tensile 
strength,  and  less  elongation  than  the  axles.  The  moduli  of 
elasticity  of  the  two  are  almost  identical. 

"  Between  loads  of  15000  and  25000  pounds  per  square  inch 
the  muck-bar  had  a  modulus  of  elasticity  of  29400000  pounds, 
the  axles  (average  of  all  specimens)  between  5000  and  20000 
pounds  per  square  inch  was  29367000  pounds.  Individually 


COLD   CRYSTALLIZATION  OF  IKON  AND   STEEL.        4/9 

the  axles  showed  the  modulus  of  elasticity  to  be  substantially 
the  same  in  each." 

Two  specimens  were  subjected  to  their  maximum  load  and 
removed  from  the  testing-machine  before  breaking  in  order  to 
see  whether  the  straining  followed  by  rest  will  cause  any 
change. 

"  It  does  not  appear  from  these  tests  that  95000  miles 
run  has  produced  any  effect  on  the  quality  of  the  metal." 

On  page  1619  et  scq.  of  the  Report  for  1886  is  given  an 
account  of  the  tests  made  on  some  more  of  these  axles  which 
had  run  163138  miles,  and  the  following  is  quoted  from  that 
account : — 

"  Specimens  from  muck-bar  axle  No.  4  after  the  axle  had 
run  163138  miles. 

"  Comparing  these  results  with  earlier  tests  of  this  series,  the 
tensile  strength  of  the  metal  in  this  axle  is  lower,  and  the 
modulus  of  elasticity  less  than  shown  by  the  preceding  axles. 

"  The  variations  in  strength,  elasticity,  and  ductility  are  no 
greater,  however,  than  those  met  in  different  specimens  of  new 
iron  of  nominally  the  same  grade,  and  while  apparently  there 
is  a  deterioration  in  quality,  it  needs  confirmation  of  a  more 
decisive  nature  from  the  remaining  axles  before  attributing 
this  result  to  the  influence  of  the  work  done  in  service." 

Another  set  of  tests  made  at  Watertown  Arsenal  is  to  be 
found  on  page  1044  et  seq.  of  the  Report  for  1885.  There  were 
tested - 

i°.  Two  side-rods  of  a  passenger  locomotive  which  had  been 
in  service  about  twelve  years. 

2°.  One  side-rod  of  a  passenger  engine  which  had  been  run 
twenty-eight  years  and  eight  months. 

3°.  One  main-rod  which  had  been  run  thirty-two  years  and 
eight  months  in  freight  and  five  years  in  passenger  service. 

In  none  of  these  tests  were  there  any  evidences  of  crystal- 
lization, as  the  metal  was  in  all  cases  fibrous  when  fractured. 


480  APPLIED  MECHANICS. 

In  the  report  is  said  :  — 

"  There  are  no  data  at  command  telling  what  the  original 
qualities  of  the  metal  of  these  bars  were  :  it  is  sufficient,  how- 
ever, to  find  toughness  and  a  fibrous  appearance  in  the  iron  to 
prove  that  brittleness  or  crystallization  has  not  resulted  from 
long  exposure  to  the  stresses  and  vibrations  these  bars  have 
sustained." 

The  only  other  evidence  that  will  be  referred  to  is  the  paper 
of  Mr.  A.  F.  Hill  upon  the  "  Crystallization  of  Iron  and  Steel," 
contained  in  the  Proceedings  of  the  Society  of  Arts  of  the 
Massachusetts  Institute  of  Technology  for  1882-83.  In  this 
article  Mr.  Hill  covers  the  ground  very  fully,  and  distinctly 
asserts  that  —  • 

"  The  fact  is  that  there  is  at  present  not  a  single  well- 
authenticated  instance  of  iron  or  steel  ever  having  become 
crystallized  from  use  under  temperatures  below  900°  F." 

He  claims  to  have  investigated  a  great  many  cases  where 
cold  crystallization  has  been  claimed,  and  to  have  found,  in 
every  case  where  crystals  existed,  that  at  some  period  of  its 
manufacture  or  working  the  metal  was  overheated.  He 
says :  — 

"  That  the  crystalline  appearance  of  a  fracture  is  not  neces- 
sarily an  indication  of  the  presence  of  genuine  crystals  is  proven 
by  the  well-known  fact  that  a  skilful  blacksmith  can  fracture 
fine  fibrous  iron  or  steel  in  such  a  manner  as  to  let  it  appear 
either  fibrous  and  silky,  or  coarse  and  crystalline,  according  to 
his  method  of  breaking  the  bar.  On  the  other  hand,  where 
there  is  genuine  crystallization,  no  skill  of  manipulation  will 
avail  to  hide  th'at  fact  in  the  fracture.  The  most  striking 
illustrations  of  this  that  have  come  under  my  notice  are  the 
fractures  of  the  beam-strap  of  the  Kaaterskill,  and  of  the 
connecting-rod  of  the  chain-cable  testing-machine  at  the  Wash- 
ington Navy  Yard.  The  photographs  of  both  fractures  are 
submitted  to  you,  and  the  similarity  of  their  appearance  is 


COLD   CRYSTALLIZATION  OF  IRON  AND   STEEL.        481 

most  singular.  Yet  what  a  difference  in  the  development  of 
the  longitudinal  sections  by  acid  treatment,  which  are  also 
presented  to  you. 

"  In  the  Kaaterskill  accident  the  fractures  of  both  the 
upper  and  lower  arms  of  the  strap  were  found  to  be  short  and 
square.  The  appearance  of  the  fractured  faces  showed  no 
trace  of  fibre,  and  was  altogether  granular.  Yet  the  longitudi- 
nal section,  taken  immediately  through  the  break,  and  devel- 
oped by  acid  treatment,  shows  the  presence  of  but  few  and 
small  crystals,  and  the  generally  fibrous  character  of  the  iron 
used  in  the  strap. 

"  In  the  connecting-rod  of  the  chain-cable  testing-machine 
we  find  the  crystalline  appearance  of  the  fracture  less,  if  any- 
thing, than  that  of  the  beam-strap,  while  the  development  of 
the  longitudinal  section  by  acid  treatment  reveals  most  beauti- 
fully, in  this  case,  the  thoroughly  crystalline  character  of  the 
metal.  As  is  w^ll  known,  this  rod,  after  many  years  of  service, 
finally  broke  under  a  comparatively  light  strain,  and  having  all 
along  been  supposed  to  have  been  carefully  made,  and  from 
well-selected  scrap,  its  intensely  crystalline  structure,  as  re- 
vealed by  the  fractures,  has  done  service  for  quite  a  number  of 
years  as  price  de  resistance  in  all  the  '  cold-crystallization ' 
arguments  which  have  been  served  up  in  that  time." 

He  then  goes  on  to  say  that  he  cut  the  rod  in  a  longitudinal 
direction,  and  treated  the  section  with  acid  ;  that  some  of  the 
crystals  shown  are  so  large  as  to  be  discernible  with  the  naked 
eye  ;  that  the  treated  section  furnished  incontrovertible  evi- 
dence that  the  rod,  aside  from  the  fact  of  being  badly  dimen- 
sioned anyhow,  was  made  of  poor  material,  badly  heated,  and 
insufficiently  hammered,  all  records,  suppositions,  and  asser- 
tions to  the  contrary  notwithstanding ;  that  there  are  a  large 
number  of  crystals  composed  of  a  substance,  presumably  a 
ferro-carbide,  which  is  not  soluble  in  nitric  acid,  and  is  found 
in  steel  only  ;  that  the  deduction  from  the  large  amount  of  this 


482  APPLIED  MECHANICS. 

substance  is  that  the  pile  was  formed  of  rather  poorly  selected 
scrap,  with  steel  scrap  mixed  in  ;  that  evidences  of  bad  heating 
are  abundant  throughout ;  and  that  the  strongest  evidence 
against  the  presumption  that  these  crystals  were  formed  during 
the  service  of  the  rod,  or  while  the  metal  was  cold,  is  found  in 
the  groupings  of  the  crystals  during  their  formation,  as  shown 
in  the  tracing  developed  by  the  acid  ;  that  they  are  not  of  the 
same  chemical  composition,  the  lighter  parts  containing  much 
more  carbon  than  the  darker  ones ;  it  is  therefore  pretty  evi- 
dent that  with  the  grouping  of  the  crystals  a  segregation  of 
like  chemical  compounds  took  place,  and  this  of  course  would 
have  been  impossible  in  the  solid  state.  He  then  cites  an 
experiment  he  made,  in  which  he  took  a  slab  of  best  selected 
scrap  weighing  about  200  pounds  and  forged  it  down  to  a 
3-inch  by  3-inch  square  bar,  one-half  being  properly  forged, 
and  the  other  half  being  exposed  to  a  sharp  flame  bringing  it 
quickly  to  a  running  heat,  keeping  it  at  this  heat  some  time, 
and  then  hammering  lightly  and  then  treating  it  a  second  time 
in  a  similar  manner  ;  the  result  being,  that  while  no  difference 
was  discernible  in  the  appearance  of  the  two  portions,  when 
cut  and  treated  with  acid  the  portion  that  was  properly  made 
showed  itself  to  be  a  fair  representative  of  the  best  quality  of 
iron,  while  in  the  other  portion  the  crystallization  was  strongly 
marked,  the  majority  of  the  crystals  being  large  and  well 
developed. 

He  also  says  :  — 

"  The  fact  is,  all  hammered  iron  or  steel  is  more  or  less 
crystalline,  the  lesser  or  greater  degree  of  crystallization  de- 
pending altogether  upon  the  greater  or  lesser  skill  employed 
in  working  the  metal,  and  also  largely  upon  the  size  of  the 
forging.  Crystallization  tends  to  lower  very  sensibly  the  elastic 
limit  of  iron  and  steel,  and  therefore  hastens  the  deterioration 
of  the  metal  under  strain.  It  is  for  this  reason  that  large  and 
heavy  forgings  ought  to  be,  and  measurably  are,  excluded  as 


EFFECT  OF   TEMPERATURE   ON  IRON  AND   STEEL.    483 

much  as  possible  from  permanent  structures.  In  machine  con- 
struction we  cannot  do  without  them,  and  must  therefore 
accept  the  necessity  of  replacing  more  or  less  frequently  the 
parts  doing  the  heaviest  work." 

The  evidence  given  above  seems  to  the  writer  to  be  suffi- 
cient, and  to  warrant  the  conclusions  stated  on  pages  475,  476. 

EFFECT  OF  TEMPERATURE  UPON  THE  RESISTING  PROPER- 
TIES OF  IRON  AND  STEEL. 

Much  the  best  and  most  systematic  work  upon  this  subject 
has  been  done  at  the  Watertown  Arsenal,  and  an  account  of  it 
is  to  be  found  in  "  Notes  on  the  Construction  of  Ordnance, 
No.  50,"  published  by  the  Ordnance  Department  at  Washing- 
ton, D.  C,  U.S.A. 

Other  references  are  the  following :  — 

Sir  William  Fairbairn:  Useful  Information  for  Engineers. 
Committee  of  Franklin  Institute:  Franklin  Institute  Journal. 
Knutt  Styffe  and  Christer  P.  Sandberg:  Iron  and  Steel. 
Kollman:  Engineering,  July  30,  1880. 
Massachusetts  R.  R.  Commissioners'  Report  of  1874. 
Bauschinger:  Mittheilungen,  Heft  13,  year  1886. 

A  summary  of  the  Watertown  tests,  largely  quoted  from 
the  above-mentioned  report,  will  be  given  here,  and  then  a  few 
remarks  will  suffice  for  the  others. 

The  subjects  upon  which  experiments  were  made  at  Water- 
town  were  the  effect  of  temperatures  upon  — 

i°.  The  coefficient  of  expansion. 

2°.  The  modulus  of  elasticity. 

3°.  The  tensile  strength. 

4°.  The  elastic  limit. 

5°.  The  stress  per  square  inch  of  ruptured  section. 

6°.  The  percentage  contraction  of  area. 

7°.  The  rate  of  flow  under  stress. 

8°.  The  specific  gravity. 


484 


APPLIED  MECHANICS. 


9°.  The  strength  when  strained  hot  and  subsequently  rup- 
tured cold. 

10°.  The  color  after  cooling. 
11°.  Riveted  joints. 

1°.    THE    COEFFICIENTS    OF    EXPANSION. 

These  were  determined  from  direct  measurements  upon  the 
experimental  bars,  first  measuring  their  lengths  on  sections 
35  inches  long,  while  the  bars  were  immersed  in  a  cold  bath  of 
ice-water,  and  again  measuring  the  same  sections  after  a  period 
of  immersion  in  a  bath  of  hot  oil. 

The  range  of  temperature  employed  was  about  210  degrees 
Fahr.,  as  shown  by  mercurial  thermometers. 

Observations  were  repeated,  and  again  after  the  steel  bars 
had  been  heated  and  quenched  in  water  and  in  oil. 

The  average  values  are  exhibited  in  the  following :  — 

TABLE  I. 

• 

First  Series  of  Bars. 


Metal. 

Chemical  Composition. 

Coefficients    of    Expansion 
per    Degree    Fahr.,    per 
Unit  of  Length. 

C. 

Mn. 

Si. 

Wrought-iron. 

.0000067302 

Steel. 

.09 

.11 

.0000067561 

a 

.20 

•45 

.0000066259 

« 

•31 

•57 

.0000065149 

«« 

•37 

.70 

.0000066597 

<  < 

•  5i 

•58 

.02 

.0000066202 

« 

•57 

•93 

.07 

.0000063891 

n 

•7i 

•58 

.08 

.0000064716 

f€ 

.81 

•56 

•17 

.0000062167 

11 

.89 

•57 

.19 

.0000062335 

«  ( 

•97 

.80 

.28 

.0000061700 

Cast-  (gun)  iron. 

.0000059261 

Drawn  copper. 

.0000091286 

EFFECT  OF   TEMPERA  TURE   ON  IRON  AND   STEEL.    485 


Subsequent  determinations  of  the  coefficient  of  expansion 
of  a  second  series  of  steel  bars  gave  — 


TABLE  II. 


Chemical  C 

omposition 

Coefficients  of  Expansion 

c. 

Mn. 

Si. 

S. 

P. 

Cu. 

per  Degree  Fahr.,  per 
Unit  of  Length. 

•  17 

I-  IS 

.023 

.122 

.079 

.04 

.0000067886 

.20 

.69 

-037 

•13 

.078 

.26 

.0000068567 

.21 

.26 

.08 

.14 

•  059 

.OO 

.0000067623 

.26 

.07 

.11 

.096 

.08 

.047 

.0000067476 

.26 

.26 

.07 

.112 

.06 

.038 

.0000067102 

.26 

.28 

.07 

•US 

.062 

•035 

.0000067175 

.28 

•23 

.09 

.168 

.09 

.I78 

.0000067794 

•43 

•97 

•05 

.08 

.096 

.024 

.0000066124 

•43 

i.  08 

•037 

.08 

.114 

•233 

.0000066377 

•53 

•75 

.10 

.078 

.087 

.174 

.0000064181 

•55 

i.  02 

•05 

.078 

.12 

•15 

.0000o66l22 

.72 

.70 

.18 

.07 

•13 

•23 

.0000064330 

.72 

•  76 

.20 

.056 

.086 

.186 

.0000063080 

•79 

.86 

.21 

.084 

•093 

.096 

.0000063562 

.07 

.07 

•13 

.01 

.018 

.006 

.0000061528 

.08 

.12 

.19 

.Oil 

.02 

trace 

.0000061702 

.12 

.10 

.09 

.013 

.018 

trace 

.0000060716 

.14 

.10 

•15 

trace 

.018 

trace 

.0000062589 

•17 

.10 

.10 

trace 

.018 

0 

.0000061332 

•31 

•13 

.19 

.Oil 

.026 

trace 

.0000061478 

Ten  bars  of  the  first  series  were  now  heated  a  bright  cherry- 
red  and  quenched  in  oil  at  80°  Fahr.,  the  hot  bars  successively 
raising  the  temperature  of  the  oil  to  about  240°  Fahr.,  the  bath 
being  cooled  between  each  immersion. 

The  behavior  of  the  bars  under  rising  temperature,  when 
examined  for  coefficients  of  expansion,  seemed  somewhat 
erratic,  the  highest  temperature  reached  being  235° ;  but  this 
behavior  was  subsequently  explained  by  the  permanent  changes 
in  length  found  when  the  bars  were  returned  to  the  cold  bath. 


486 


APPLIED  MECHANICS. 


Generally  the  bars  were  found  permanently  shortened  at  the 
close  of  these  observations. 

The  bars  were  again  heated  bright  cherry-red  and  quenched 
in  water  at  50°  to  55°  Fahr.,  the  water  being  raised  by  the 
quenching  to  110°  to  125°  Fahr. 

After  resting  72  hours,  measurements  were  taken  in  the  cold 
bath,  followed  by  a  rest  of  18  hours,  when  they  were  heated 
and  measure^  in  the  hot  bath,  after  which  they  were  measured 
in  the  cold  bath ;  the  maximum  temperature  reached  with  the 
hot  bath  being  233°. 7  Fahr.,  erratic  behavior  occurring  still. 

They  were  next  heated  in  an  oil  bath  at  300°  Fahr.,  and 
kept  at  this  temperature  6  hours,  then  cooled  in  the  bath  ;  15 
hours  later  they  were  heated  to  243°  Fahr.,  and  again  measured 
hot,  and  then  cold.  These  downward  readings  showed  the 
quenched  in  water  bars  to  have  their  coefficients  elevated 
above  the  normal,  as  shown  in  the  following  table,  these 
being  the  same  steel  bars  as  in  Table  I,  and  in  the  same 
order :  — 

TABLE    III. 


Coefficients  of    Expansion 

Apparent  Shortening  of  Bars 
Due  to  Six   Hours  at   300° 

per  Degree  per  Unit  of 

.Fahr.,    and    the   following 

Length. 

Immersion  in  the  Hot  Bath. 

.0000067641 

—  .0006 

.0000066622 

.0002 

.0000066985 

.0016 

.0000067377 

.0023 

'     .0000069776 

—  .0004 

.000006704! 

.0082 

.0000066939 

.0064 

.0000068790 

.0054 

.0000072906 

•0055 

.0000071578 

.0048 

EFFECT  OF   TEMPERATURE   ON  IRON  AND   STEEL,    487 

Finally  the  bars  were  annealed  by  heating  bright  red  and 
cooling  in  pine  shavings,  the  effect  of  which  was  to  approxi- 
mately restore  the  rate  of  expansion  to  the  normal,  as  shown 
by  Table  I  for  these  ranges  of  temperature. 


These  were  obtained  with  the  first  series  of  bars  at  atmos- 
pheric temperatures,  and  at  higher  temperatures,  up  to  495° 
Fahr. 

There  occurred  invariably  a  decrease  in  the  modulus  of 
elasticity  with  an  increase  in  temperature,  and,  in  the  case  of 
the  specimens  tested,  the  low  carbon  steels  showed  a  greater 
reduction  in  the  modulus  than  the  high  carbon  steels,  the 
first  specimen  having  a  modulus  of  elasticity  at  the  minimum 
temperature  30612000,  and  at  the  maximum  27419000,  while 
the  last  specimen  had  at  the  minimum  temperature  29126000, 
and  at  the  maximum  27778000. 

3°.    TENSILE    STRENGTH. 

The  tests  were  made  upon  the  first  series  of  steel  bars, 
wrought-irons  marked  A  and  B,  a  muck-bar  railway  axle,  and 
cast-iron  specimens  from  a  slab  of  gun-iron. 

The  specimens  were  o".798  diameter,  and  5"  length  of  stem, 
having  threaded  ends  i' '.25  diameter. 

Wrought-iron  A  was  selected  because  it  was  found  very  hot 
short  at  a  welding  temperature.  It  had  been  strained  with  a 
tensile  stress  of  42320  pounds  per  square  inch  seven  years 
previous  to  being  cut  up  into  specimens  for  the  hot  tests. 

The  specimens  while  under  test  were  confined  within  a 
sheet-iron  muffle,  through  the  ends  of  which  passed  auxiliary 
bars  screwed  to  the  specimens,  the  auxiliary  bars  being  secured 
to  the  testing-machine. 


488  APPLIED  MECHANICS. 

The  heating  was  done  by  means  of  gas-burners  arranged 
below  the  specimen  and  within  the  muffle. 

The  temperature  of  the  test-bar  was. estimated  from  the 
expansion  of  the  metal,  observed  on  a  specimen  length  of  six 
inches,  using  the  coefficients  which  were  determined  at  lower 
temperatures,  as  hereinbefore  stated,  assuming  there  was  a 
uniform  rate  of  expansion. 

Access  to  the  specimen  for  the  purpose  of  measuring  the 
expansion  was  had  through  holes  in  the  top  of  the  muffle. 
The  temperature  was  regulated  by  varying  the  number  of  gas- 
burners  in  use,  the  pressure  of  the  gas,  and  also  by  means  of 
diaphragms  placed  within  the  muffle  for  diffusing  the  heat. 

The  approximate  elongations  under  different  stresses  were 
determined  during  the  continuance  of  a  test  from  measurements 
made  on  the  hydraulic  holders  of  the  testing-machine,  at  a 
convenient  distance  from  the  hot  muffle,  correcting  these 
measurements  from  data  obtained  by  simultaneous  micrometer 
readings  made  on  the  specimen  and  the  hydraulic  holders  at 
atmospheric  temperatures. 

While  it  does  not  seem  expedient  in  one  series  of  tests  to 
obtain  complete  results  upon  the  tensile  properties  at  high 
temperatures,  yet,  incidentally,  much  additional  valuable  infor- 
mation may  be  obtained  while  giving  prominence  to  one  or 
more  features. 

From  these  elongations  the  elastic  limits  were  established 
where  the  elongations  increased  rapidly  under  equal  incre- 
ments of  load.  Proceeding  with  the  test  until  the  maximum 
stress  was  reached,  recorded  as  the  tensile  strength,  observing 
the  elongation  at  the  time,  then,  when  practicable,  noting 
the  stress  at  the  time  of  rupture." 

For  the  detailed  tables  of  tests  the  student  is  referred  to 
the  "  Notes  on  the  Construction  of  Ordnance." 

The  elastic  limits  and  tensile  strengths  are  computed  in 
pounds  per  square  inch,  both  on  original  sectional  areas  of  the 


EFFECT  OF  TEMPERATURE   ON  IRON  AND   STEEL.    489 

specimens  and  on  the  minimum  or  reduced  sections,  as  meas- 
ured at  the  close  of  the  hot  tests. 

From  the  results  it  appears  that  the  tensile  strength  of  the 
steel  bars  diminishes  as  the  temperature  increases  from  zero 
Fahr.,  until  a  minimum  is  reached  between  200°  and  300°  Fahr., 
the  milder  steels  appearing  to  reach  the  place  of  minimum 
strength  at  lower  temperatures  than  the  higher  carbon  bars. 

From  the  temperature  of  this  first  minimum  strength  the 
bars  display  greater  tenacity  with  increase  of  temperature,  until 
the  maximum  is  reached  between  the  temperatures  of  about 
400°  to  650°  Fahr. 

The  higher  carbon  steels  reach  the  temperature  of  maximum 
strength  abruptly,  and  retain  the  highest  strength  over  a  lim- 
ited range  of  temperature.  The  mild  steels  retain  the  increased 
tenacity  over  a  wider  range  of  temperature. 

From  the  temperature  of  maximum  strength  the  tenacity 
diminishes  rapidly  with  the  high  carbon  bars,  somewhat  less 
,so  with  mild  steels,  until  the  highest  temperatures  are  reached, 
covered  by  these  experiments. 

The  greatest  loss  observed  in  passing  from  70°  Fahr.  to  the 
temperature  of  first  minimum  strength  was  6.5  per  cent  at 
295°  Fahr. 

The  greatest  gain  over  the  strength  of  the  metal  at  70°  was 
25.8  per  cent  at  460°  Fahr. 

The  several  grades  of  metal  approached  each  other  in 
tenacity  as  the  higher  temperatures  were  reached.  Thus  steels 
differing  in  tensile  strength  nearly  90000  pounds  per  square 
inch  at  70°,  when  heated  to  1600°  Fahr.  appear  to  differ  only 
about  loooo  pounds  per  square  inch. 

The  rate  of  speed  of  testing  which  may  modify  somewhat 
the  results  with  ductile  material  at  atmospheric  temperatures 
has  a  very  decided  influence  on  the  apparent  tenacity  at  high 
temperatures. 

A  grade   of   metal  which,  at  low  temperatures,  had  little 


49°  APPLIED   MECHANICS. 

ductility,  displayed  the  same  strength  whether  rapidly  or  slowly 
fractured  from  the  temperature  of  the  testing-room  up  to  600° 
Fahr. ;  above  this  temperature  the  apparent  strength  of  the 
rapidly  fractured  specimens  largely  exceeded  the  others. 

At  1410°  Fahr.  the  slowly  fractured  bar  showed  33240 
pounds  per  square  inch  tensile  strength,  while  a  bar  tested  in 
two  seconds  showed  63000  pounds  per  square  inch. 

Cast-iron  appeared  to  maintain  its  strength  with  a  tendency 
to  increase  until  900°  Fahr.  is  reached,  beyond  which  the 
strength  diminishes.  Under  the  higher  temperatures  it  devel- 
oped numerous  cracks  on  the  surface  of  the  specimens  preced- 
ing complete  rupture. 

4°.    ELASTIC    LIMIT. 

The  report  says  of  this  that  it  appears  to  diminish  with  in- 
crease of  temperature.  Owing  to  a  period  of  rapid  yielding  with- 
out increase  of  stress,  or  even  under  reduced  stress,  the  elastic 
limit  is  well  defined  at  moderate  temperatures  with  most  of  the 
steels. 

Mild  steel  shows  this  yielding  point  up  to  the  vicinity  of 
500°;  in  hard  steels,  if  present,  it  appears  at  lower  temperatures. 

The  gradual  change  in  the  rate  of  elongation  at  other  times 
often  leaves  the  definition  of  the  elastic  limit  vague  and  doubt- 
ful, especially  so  at  high  temperatures.  The  exclusion  of  de- 
terminable  sets  would  in  most  cases  place  the  elastic  limit  below 
the  values  herein  given. 

In  approaching  temperatures  at  which  the  tensile  proper- 
ties are  almost  eliminated  exact  determinations  are  correspond- 
ingly difficult,  the  tendency  being  to  appear  to  reach  too  high 
values. 

5°.    STRESS   ON    THE    RUPTURED    SECTION. 

This,  generally,  follows  with  and  resembles  the  curve  of 
tensile  strength. 


EFFECT  OF   TEMPERATURE    ON  IRON  AND   STEEL.    49 1 

Specimens  of  large  contraction  of  area,  tested  at  high 
temperature,  have  given  evidence  on  the  fractured  ends  of 
having  separated  at  the  centre  of  the  bar  before  the  outside 
metal  parted. 

Elongation  under  Stress* 

Although  the  metal  is  capable  of  being  worked  under  the 
hammer  at  high  temperatures,  it  does  not  then  possess  sufficient 
strength  within  itself  to  develop  much  elongation,  general 
elongation  being  greatest  at  lower  temperatures. 

Greater  rigidity  exists  under  certain  stresses  at  intermedi- 
ate temperatures  than  at  either  higher  or  lower  temperatures. 

Thus  one  of  the  specimens  tested  at  569°  Fahr.  showed  less 
elongation  under  stresses  above  50000  pounds  per  square  inch 
than  the  bars  strained  at  higher  or  lower  temperatures. 

Two  other  specimens  showed  a  similar  behavior  at  315°  and 
387°  respectively,  and  likewise  other  specimens. 

In  bars  tested  at  about  200°  to  400°  Fahr.  there  are  dis- 
played alternate  periods  of  rigidity  and  relaxation  under  in- 
creasing stresses,  resembling  the  yielding  described  as  occur- 
ring with  some  bars  immediately  after  passing  the  elastic 
limit. 

The  repetition  of  these  intervals  of  rigidity  and  relaxation 
is  suggestive  of  some  remarkable  change  taking  place  within 
the  metal  in  this  zone  of  temperature. 

6°.    PERCENTAGE    CONTRACTION^  OF   AREA. 

This  varies  with  the  temperature  of  the  bar ;  it  is  somewhat 
less  in  mild  and  medium  hard  steels  at  400°  to  600°  than  at 
atmospheric  temperatures. 

Above  500°  or  600°  the  contraction  increases  .with  the 
temperature  of  the  metal ;  with  three  exceptions,  which  showed 
diminished  contraction  at  1100°  Fahr.,  until  at  the  highest 
temperatures  some  of  them  were  drawn  down  almost  to  points. 


492  APPLIED  MECHANICS. 

7°.    RATE    OF    FLOW    UNDER   STRESS. 

The  full  effect  of  a  load  superior  to  the  elastic  limit  is  not 
immediately  felt  in  the  elongation  of  a  ductile  metal,  and  the 
same  is  true  at  higher  temperatures. 

The  flow  caused  by  a  stress  not  largely  in  excess  of  the 
'elastic  limit  has  a  retarding  rate  of  speed,  and  eventually  ceases 
altogether  ;  whereas  under  a  high  stress  the  rate  of  flow  may 
accelerate,  and  end  in  rupture  of  the  metal. 

Hence  the  apparent  tensile  strength  maybe  modified  within 
limits  by  the  time  employed  in  producing  fracture. 

8°.    SPECIFIC    GRAVITY. 

In  general,  the  specific  gravity  is  materially  diminished  in 
the  vicinity  of  the  fractured  ends  of  tensile  specimens,  and  this 
diminution  takes  place  in  the  different  grades  of  steel,  in  bars 
ruptured  under  different  conditions  of  temperature,  stress,  and 
contraction  of  area. 

9°.    BARS    STRAINED    HOT,    AND    SUBSEQUENTLY    RUPTURED    COLD. 

The  effect  of  straining  hot  on  the  subsequent  strength  cold 
appears  to  depend  upon  the  magnitude  of  the  straining  force 
and  the  temperature  in  the  first  instance. 

There  is  a  zone  of  temperature  in  which  the  effect  of  hot 
straining  elevates  the.elastic  limit  above  the  applied  stress,  and 
above  the  primitive  value,  and  if  the  straining  force  approaches 
the  tensile  strength,  there  is  also  a  material  elevation  of  that 
value  when  ruptured  cold.  These  effects  have  been  observed 
within  the  limits  of  about  335°  and  740°  Fahr. 

After  exposure  to  higher  temperatures  there  occurs  a 
gradual  loss  in  both  the  elastic  limit  and  tensile  strength,  and 
generally  a  noticeable  increase  in  the  contraction  of  area. 


EFFECT  OF  TEMPERATURE   ON  IRON  AND   STEEL.    49$ 


10  .    COLOR   AFTER    COOLING. 

This  was  not  sensibly  changed  by  temperatures  below  200°. 
After  300°  the  metal  was  light  straw-colored  ;  after  40x3°,  deep 
straw ;  from  500°  to  600°,  purple,  bronze-colored,  and  blue ; 
after  700°,  dark  blue  and  blue  black. 

After  800°  some  specimens  still  remained  dark  blue.  After 
heating  above  about  800°  the  final  color  affords  less  satisfactory 
means  of  approximately  judging  of  the  temperature,  the  color 
remaining  a  blue  black,  and  darker  when  a  thick  magnetic 
oxide  is  formed. 

At  about  1100°  the  surface  oxide  reaches  a  tangible  thick- 
ness, a  heavy  scale  of  o".ooi  to  o".oo2  thickness  forming  as 
higher  temperatures  are  reached.  The  red  oxide  appears  at 
about  1500°. 

11°.    IN    THE   TESTS   OF   RIVETED   JOINTS 

of  steel  boiler-plates  at  temperatures  ranging  from  70°  to  about 
700°  Fahr.  the  indications  of  the  tensile  tests  of  plain  bars  were 
corroborated. 

Joints  at  200°  Fahr.  showed  less  strength  than  when  cold ; 
at  250°  and  higher  temperatures  the  strength  exceeded  the 
cold  joints ;  and  when  overstrained  at  400°  and  500°  there  was 
found,  upon  completing  the  test  cold,  an  increase  in  strength. 

Rivets  which  sheared  cold  at  40000  to  41000  pounds  per 
square  inch,  at  300°  Fahr.  sheared  at  46000  pounds  per  square 
inch ;  and  at  600°  Fahr.,  the  highest  temperature  at  which  the 
joints  failed  in  this  manner,  the  shearing-strength  was  42130 
pounds  per  square  inch. 

In  addition  to  the  work  at  Watertown  which  has  just  been 
detailed  two  other  matters  will  be  referred  to  here. 


494  APPLIED  MECHANICS. 

1°.  It  is  well  known  that  wrought-iron  and  steel  are  very 
brittle  at  a  straw  heat  and  a  pale  blue,  as  shown  by  the  fact  that 
when  the  attempt  is  made  to  bend  a  specimen  at  these  tempera- 
tures it  results  in  cracking  it  some  time  before  a  complete  bend- 
ing can  be  effected,  even  in  the  case  of  metal  which  is  so  ductile 
that  it  can  be  bent  double  cold,  red  hot,  or  at  a  flanging  heat, 
without  showing  any  signs  of  cracking. 

2°.  Bauschinger  defines  the  elastic  limit  as  the  load  at  which 
the  stress  is  no  longer  proportional  to  the  strain ;  whereas  he 
calls  stretch-limit  (Streckgrenze)  the  load  at  which  the  strain 
diagram  makes  a  sudden  change  in  its  direction  ;  i.e.,  where 
instead  of  showing  a  gradually  increasing  ratio  of  strain  to  stress 
it  shows  a  sudden  and  rapid  increase. 

From  his  experiments  (see  Heft  13  of  the  Mittheilungen, 
year  1886)  he  draws  the  following  conclusions  :  — 

(a)  That  the  effect  of  heating  and  subsequent  cooling  in 
lowering  both  the  elastic  and  the  stretch  limits  in  mild   steel 
begins  at  about  660°  Fahr.  when  the  cooling  is  sudden,  and  at 
about  840°  Fahr.  when  it  is  slow,  and  for  wrought-iron  at  about 
750°  with  either  rapid  or  slow  cooling. 

(b)  That  the  operation  of  heating  above  those  temperatures, 
and  of  subsequent  slow  or  quick  cooling,  is  that  both  the  elastic 
and  the  stretch  limit  are  lowered,  and  the  more  so  the  greater 
the  heating ;  also,  that  this  effect  is  greater  on  the  elastic  than 
on  the  stretch  limit. 

(c)  Quick  cooling  after  heating  higher  than  the  above-stated 
temperatures  lowers  the  elastic  and  the  stretch  limit,  especially 
the  first,  much  more  than  slow  cooling,  dropping  the  elastic 
limit  almost  immediately  at  a  heat  of  about  930°  and  certainly 
at  a  red  heat  to  aothing  or  nearly  nothing  in  wrought-iron,  and 
in  both  mild  and  hard  steel,  while  slow  cooling  cannot  bring 
about  such  a  great  drop  of  the  elastic  limit,  even  from  more 
than  a  red  heat. 


COLD   CRYSTALLIZATION  OF  IRON  AND   STEEL.        495 


Effect  of  Cold-Rolling  on  Iron  and  Steel. — It  has  already 
been  stated,  p.  410,  that  it  was  discovered  independently  by 
Commander  Beardslee  and  Professor  Thurston,  that  if  a  load 
were  gradually  applied  to  a  piece  of  iron  or  steel  which  exceeded 
its  elastic  limit,  and  the  piece  then  allowed  to  rest,  the  elastic 
limit  and  the  ultimate  strength  would  thus  be  increased.  This 
may  be  accomplished  with  soft  iron  and  steel  by  cold-rolling  or 
cold-drawing,  but  cannot  be  taken  advantage  of  in  hard  iron 
or  steel. 

«  Professor  Thurston,  who  has  investigated  this  matter  at 
great  length,  and  made  a  large  number  of  tests  on  the  subject, 
gives  the  following  as  the  results  of  cold-rolling:  — 


Increase  in 

Per  Cent. 

2<J  to     4O 

Transverse  stress           

50  to    80 

Elastic  limit  (tension,  torsion,  and  transverse), 

80  to  125 

-JQO  to  400 

Elastic  resilience  (transverse) 

1  1O  to  42  ^ 

He  also  says,  in  regard  to  the  modulus  of  elasticity,  — 

"  Collating  the  results  of  several  hundred  tests,  the  author 
[Professor  Thurston]  found  that  the  modulus  of  elasticity  rose, 
in  cold-rolling,  from  about  25000000  Ibs.  per  square  inch  to 
26000000,  the  tenacity  from  52000  Ibs.  to  nearly  70000,  the 
elastic  limit  from  30000  Ibs.  to  nearly  60000  Ibs.  ;  and  the  ex- 
tension was  reduced  from  25  to  loj  per  cent. 

"  Transverse  loads  gave  a  reduction  of  the  modulus  of  elas- 
ticity to  the  extent  of  about  1000000  Ibs.  per  square  inch,  an 
increase  in  the  modulus  of  rupture  from  73600  to  133600,  and 
reduction  of  deflection  at  maximum  load  of  about  25  per  cent. 
The  resistance  of  the  elastic  limit  was  doubled,  and  occurred 
at  a  much  greater  deflection  than  with  untreated  iron." 

On  the  other  hand,  the  two  steel  eye-bars  referred  to  on 


496 


APPLIED  MECHANICS. 


p.  472  show  a  decrease  of  modulus  of  elasticity  with  increasing 
over-strain. 

Whitworttis  Compressed  Steel. — Sir  Joseph  Whitworth  pro- 
duces steel  of  great  strength  by  applying  to  the  molten  metal,, 
directly  after  it  leaves  the  furnace,  a  pressure  of  about  14000. 
Ibs.  per  square  inch ;  this  being  sufficient  to  reduce  the  length 
of  an  eight-foot  column  by  one  foot.  He  claims,  according  tc* 
D.  K.  Clark,  to  be  able  to  obtain  with  certainty  a  strength  of 
40  English  tons  with  30  per  cent  ductility,  and  mild  steel  of  a 
strength  of  30  English  tons  with  33  or  34  per  cent  ductility.- 

The  following  table  is  taken  from  D.  K.  Clark's  "Rules, 
and  Tables  : "  — 


Ultimate  Tensile 

Strength,  in  Ibs., 

Elongation, 

per  Square  Inch. 

per  cent. 

Axles,  boilers,  connecting-rods,  rivets,  railway  tires,  ) 
guns,  and  gun-carriages,                                           > 

89600 

32 

Cylinder  linings,  parts  of  large  machines,  hoops,  ) 
and  trunnions,                                                            ) 

107520 

24 

Large    planing   and  lathe   tools,  shears,   smiths'  "I 

punches,  dies  and  sets,  cold-chisels,  screw  tools,  \ 

129920 

17 

etc,                                                                       J 

Boring-tools,  finishing-tools  for  planing  and  turn-  ") 

ing,                                                                              ) 

152320 

IO 

Alloyed  with  tungsten     

161280 

14. 

±  *T 

The  following  tests  were  made  on  the  Watertown  machine, 
upon  some  specimens  of  Whitworth  steel  taken  from  a  section 
of  a  jacket  which  was  shrunk  upon  a  wrought-iron  tube,  and 
removed  from  shrinkage  by  the  application  of  high  furnace 
heat:  — 


FACTOR    OF  SAFETY. 


497 


TENSILE  TESTS. 


Diameter, 
Inches. 

Tensile 
Strength, 
Ibs.  per  Sq.  In. 

Elastic  Limit, 
Ibs.  per  Sq.  In. 

Contraction 
of  Area, 
per  cent. 

0.564 

103960 

55000 

41.9 

0.564 

90040 

48000 

47.2 

0.564 

104200 

57000 

24.6 

0.564 

IOOI20 

57000 

44.6 

0.564 

93040 

53000 

39-2 

0.564 

104160 

60000 

24.6 

0.564 

93160 

47000 

39-2 

COMPRESSIVE   TESTS. 


Length, 
Inches. 

Diameter, 
Inches. 

Compressive 
Strength; 
Ibs.  per  Sq.  In. 

Elastic  Limit, 
Ibs.  per  Sq.  In. 

5 

0.798 

IO2IOO 

6lOOO 

5 

0.798 

SgOOO 

57000 

3-94 

0.798 

IOI600 

53000 

3-94 

0.798 

IOI600 

54000 

§  227.  Factor  of  Safety.  —  In  order  to  determine  the 
proper  dimensions  of  any  loaded  piece,  it  becomes  necessary 
to  fix,  in  some  way,  upon  the  greatest  allowable  stress  per 
square  inch  to  which  the  piece  shall  be  subjected. 

The  most  common  practice  has  been  to  make  this  some 
fraction  of  the  breaking-strength  of  the  material  per  square 
inch.  v 

As  to  how  great  this  factor  should  be,  depends  upon  — 

i°.  The  use  to  which  the  piece  is  to  be  subjected ; 

2°.  The  liability  to  variation  in  the  quality  of  the  material  ; 

3°.  The  question  whether  we  are  considering,  as  the  load 
upon  the  piece,  the  average  load,  or  the  greatest  load  that  can 
by  any  possibility  come  upon  it"; 

4°.  The  question  as  to  whether  the  structure  is  a  temporary 
or  a  permanent  one; 


OF  THE 

I7BES 


ITT 


498  APPLIED   MECHANICS. 

5°.  The  amount  of  injury  that  would  be  done  by  breakage 
of  the  piece ; 
and  other  considerations. 

The  factors  most  commonly  recommended  are,  3  for  a  dead 
or  quiescent  load,  and  6  for  a  live  or  moving  load. 

A  common  American  and  English  practice  for  iron  bridges 
is  to  use  a  factor  of  safety  of  4  for  both  dead  and  .moving  load. 
In  machinery  a  factor  as  large  as  6  is  desirable  when  there  is 
no  liability  to  shocks  ;  and  when  there  is,  a  larger  factor  should 
be  used. 

A  method  sometimes  followed  for  tension  and  compression 
pieces  is,  to  prescribe  that  the  stretch  under  the  given  load 
should  not  exceed  a  certain  fixed  fraction  of  the  length.  This 
requires  a  knowledge  of  the  modulus  of  elasticity  of  the  mate- 
rial. 

In  the  case  of  a  piece  subjected  to  a  transverse  load,  it  is 
the  most  common  custom  to  determine  its  dimensions  in  accord- 
ance with  the  principle  of  providing  sufficient  strength  ;  and 
for  this  purpose  a  certain  fraction  (as  one-fourth)  of  the  mod- 
ulus of  rupture  is  prescribed  as  the  greatest  allowable  safe 
stress  per  square  inch  at  the  outside  fibre.  Thus,  for  wrought- 
iron  from  10000  to  12000  Ibs.  per  square  inch  is  often  adopted 
as  the  greatest  allowable  stress  at  the  outside  fibre,  this  being 
about  one-fourth  of  the  modulus  of  rupture. 

The  other  method  for  dimensioning  a  beam  is,  to  prescribe 
its  stiffness ;  i.e.,  that  it  shall  not  deflect  under  its  load  more 
than  a  certain  fraction  of  the  span.  This  fraction  is  taken  as 
dro  to  yfo.  \ 

This  latter  method  depends  upon  the  modulus  of  elasticity 
of  the  beam ;  and  while  it  is  the  most  advisable  method  to 
follow,  and  as  a  rule  would  be  safer  than  the  other  method, 
nevertheless,  in  the  case  of  very  stiff  and  brittle  material  it 
might  be  dangerous ;  hence  we  ought  to  know  also  the  break- 
ing-weight and  the  limit  of  elasticity  of  the  beam  we  are  to  use, 


WOHLER' S  RESULTS.  499 


and  not  allow  it  to  approach  either  of  these.  This  precaution 
will  be  especially  important  to  observe  in  the  case  of  steel 
beams,  which  are  only  now  being  introduced. 

On  the  other  hand,  in  moving  machinery  a  factor  of  safety 
of  six  is  usually  required  when  there  is  no  unusual  exposure  to 
shocks,  as  in  smooth-running  shafting,  etc.  ;  and  when  there 
are  irregular  shocks  liable  to  come  upon  the  piece,  a  greater 
factor  is  used. 

WOHLER'S  RESULTS. 

§  228.  Repeated  Stresses. — The  extensive  experiments  of 
Wohler  for  the  Prussian  government,  which  were  subsequently 
carried  on  by  his  successor,  Spangenberg,  were  made  to  deter- 
mine the  effect  of  oft-repeated  stresses,  and  of  changes  of 
stress,  upon  wrought-iron  and  steel. 

In  the  ordinary  American  and  English  practice,  it  is  cus- 
tomary, in  determining  the  dimensions  of  a  piece,  as  of  abridge 
member,  to  ascertain  the  greatest  load  which  the  piece  can 
ever  be  called  upon  to  bear,  and  to  fix  the  size  of  the  piece  in 
accordance  with  this  greatest  load. 

Wohler  called  attention  to  the  fact  that  the  load  that  would 
break  a  piece  depends  upon  both  the  greatest  and  least  load 
that  it  would  ever  be  called  upon  to  bear.  Thus,  a  tension-rod 
which  is  subjected  to  alternate  changes  of  load  extending  from 
.20000  to  80000  Ibs.  would  require  a  greater  area  for  safety  than 
one  which  was  subjected  to  loads  varying  only  between  the 
limits  of  60000  and  80000  Ibs.  ;  and  this  would  require  more 
area  than  one  which  was  subjected  to  a  steady  load  of  80000 
Ibs. 

Wohler  expresses  this  law  as  follows,  in  his  "Festigkeits 
versuche  mit  Eisen  und  Stahl." 

"  The  "law  discovered  by  me,  whose  universal  application 
for  iron  and  steel  has  been  proved  by  these  experiments,  is  as 
follows :  The  fracture  of  the  material  can  be  effected  by 
variations  of  stress  repeated  a  great  number  of  times,  of 


5OO  APPLIED   MECHANICS. 

which  none  reaches  the  breaking-limit.  The  differences  of 
the  stresses  which  limit  the  variations  of  stress  determine  the 
breaking-strength.  The  absolute  magnitude  of  the  limiting 
stresses  is  only  so  far  of  influence  as,  with  an  increasing  stress, 
the  differences  which  bring  about  fracture  grow  less. 

"For  cases  where  the  fibre  passes  from  tension  to  compres- 
sion and  vice  versa,  we  consider  tensile  strength  as  positive 
and  compressive  strength  as  negative ;  so  that  in  this  case  the 
difference  of  the  extreme  fibre  stresses  is  equal  to  the  greatest 
tension  plus  the  greatest  compression." 

Besides  the  ordinary  tests  of  tensile,  compressive,  shearing, 
and  torsional  strength,  he  made  his  experiments  mainly  on  the 
following  two  cases  :  — 

1°.  Repeated  tensile  strength;  the  load  being  applied  and 
wholly  removed  successively,  and  the  number  of  repetitions 
required  for  fracture  counted. 

2°.  Alternate  tension  and  compression  of  equal  amounts 
successively  applied,  the  number  of  repetitions  required  for 
fracture  being  counted. 

In  making  these  two  sets  of  tests,  he  made  the  first  set  in 
two  ways  :  — 

(a)  By  applying  direct  tension. 

(b)  By  applying   a   transverse   load,    and   determining   the 
greatest  fibre  stress. 

The  second  set  of  tests  was  made  by  loading  at  one  end  a 
piece  of  shaft  fixed  in  direction  at  the  other,  and  then  causing 
it  to  revolve  rapidly,  each  fibre  passing  alternately  from  tension 
to  an  equal  compression,  and  vice  versa. 

He  also  tried  a  few  experiments  where  the  lower  limit  of 
stress  was  neither  zero  nor  equal  to  the  upper  limit,  with  a 
minus  sign,  also  some  experiments  on  torsion,  on  shearing, 
and  on  repeated  torsion. 

When  Wohler  had  made  his  experiments,  and  published  his 
results,  there  were  a  number  of  attempts  made  by  different 


LAUNHARDT'S  FORMULA.  50! 

persons  to  deduce  formulae  which  should  depend  upon  these 
experiments  for  their  constants,  and  which  should  serve  to 
determine  the  breaking-strength  for  any  given  variation  of 
stresses. 

Only  two  of  these  formulae  will  be  given  here,  viz.,  — 
i°.  That  of  Launhardt  for  one  kind  of  stress, 
2°.  That  of  Weyrauch  for  alternate  tension  and  compression, 
as  these  are  the  most  used  of  the  formulae  developed. 


LAUNHARDT'S   FORMULA. 

The  constants  used  in  this  formula  are  — 

i°.  t,  the  carrying-strength  of  the  material  per  unit  of  area; 
this  being  the  greatest  load  per  square  inch  of  which  the  piece 
can  bear  an  unlimited  application  without  breaking.  Practi- 
cally it  is  the  breaking-strength  per  unit  of  area. 

2°.  «,  the  primitive  safe  strength  ;  i.e.,  the  greatest  stress 
per  unit  of  area  of  which  the  piece  can  bear,  without  breaking, 
an  unlimited  number  of  repetitions,  the  load  being  entirely 
removed  between  times.  It  is  not  a  safe  but  a  breaking 
strength.  These  two  quantities  have  been  determined  experi- 
mentally by  Wb'hler ;  and  it  is  the  object  of  Launhardt's  formula 
to  deduce,  in  terms  of  /,  «,  and  the  ratio  between  the  greatest 
and  least  loads  to  which  the  piece  is  ever  subjected,  the  value 
a  of  the  breaking-strength  per  unit  of  area  when  these  loads 
are  applied. 

The  formula  and  its  deduction  are  as  follows  :  — 

Let  the  greatest  stress  per  unit  area  be  a. 
the  least  stress  per  unit  area  be  c. 
their  difference  be  d  =  a  —  c. 

Now  a  depends  on  d  for  its  value ;  and  hence  we  may  write 

a  =  ad, 
where  a  is  a  function  of  a  also. 


5O2  APPLIED   MECHANICS. 

Now,  the  two  cases  experimented  upon,  viz.,  — 

i°.  When  a  =  /,  c  =  f,  d  —  o, 

2°.  When  a  =  u,  c  —  o,  d  —  u^ 

must  be  satisfied  by  the  value  of  a  used  :  otherwise  the  value 
we  use  is  wrong,  as  these  are  two  particular  cases.  Now,  Laun- 
hardt  takes 

t  —  u 
a  =  7^' 
and  hence 


and  it  will  become  evident  that  this  value  of  a  does  satisfy  the 
two  conditions  stated  if  we  observe,  that,  in  the  first  case,  since 


we  must  have  a  —  oo  in  order  that  we  may  have 

a  =  // 
and  if  we  put  a  =  t  in  the  value  of  a,  we  obtain 

a  =  oo  : 

also  in  the  second  case,  since  d  —  u  and  a  =  u,  we  must  have 

a  =  i  • 
and  if  we  put  a  =  u  in  the  value  of  a,  we  obtain 

a  =   I. 

Hence  in  using  (i)  we  are  using  a  formula  for  a  which 
satisfies  the  two  cases  of  carrying-strength  and  of  primitive 
safe  strength  ;  and  the  question  as  to  its  being  a  suitable  value 
to  use  must  depend  upon  how  well  it  will  satisfy  intermediate 
values,  i.e.,  cases  where  the  two  extremes  are  not  those  of  the 
carrying-strength  nor  of  the  primitive  safe  strength. 


LAUNHARDT'S  FORMULA.  503 

In  the  few  cases  of  intermediate  values  experimented  upon 
by  Wohler,  there  is  a  very  close  agreement  between  the  experi- 
mental results  and  those  obtained  by  the  formula. 
.    Now,  put  d  =  a  —  c  in  (i),  and  we  have 


(2) 


at  —  a2  =  at  —  au  —  c(t  —  u) 
.'.     a2  =  au  +  (t  —  u)c 


(3) 


and  if   we   denote  by  max  L  the  greatest  load  on  the  entire 
piece,  and  by  min  L  the  least,  we  shall  have 


c       mn 


a       max  L 
Hence 


this  being  in  such  a  form  as  can  be  used.     Or  we  may  write  it 
thus: 


!t  —  u  mm  L  \ 
i  +  -  rij  (5) 

u      maxZ)'  *} 


this  being  the  more  common  form. 

The  values  of  the  constants  as  determined  by  Wohler's 
experiments,  and  the  resulting  form  of  the  formula  for  Phoenix 
axle-iron  and  for  Krupp  cast-steel,  have  already  been  given  in 
§  172. 

In  the  same  paragraph  are  given  the  corresponding  values 
of  by  the  safe  working-strength,  the  factor  of  safety  being  three. 


504  APPLIED   MECHANICS. 


WEYRAUCH'S    FORMULA    FOR    ALTERNATE    TENSION    AND 
COMPRESSION. 

The  constants  used  in  this  formula  are  :  — 

i°.  «,  the  primitive  safe  strength,  which  has  been  already 
defined. 

2°.  s,  the  vibration  safe  strength  ;  i.e.,  the  greatest  stress 
per  unit  of  area  of  which  the  piece  can  bear,  without  breaking, 
an  unlimited  number  of  applications^  the  other  extreme  stress 
which  it  bears  alternately  being  —  s. 

He  lets  a  =  greatest  stress  per  unit  of  area,  c  =  greatest 
stress  of  the  opposite  kind  per  unit  of  area.  If  a  is  tension, 
c  is  compression,  and  vice  versa. 

Then,  if  d  is  the  difference, 

d  —  a  -f  c. 
Weyrauch  writes,  as  before, 

a  =  a.d, 

and  gives  to  a  a  value  which  will  satisfy  the  two  special  cases 
experimented  upon  ;  viz.,  — 

i°.  When  a  =  u,  c  —  o,  d  =  u. 

2°.  When  a  =  s,  c  =  s,  d  =  2s. 

He  writes 

u  —  s 


...     a=  ~         d.  (6) 

2.u  —  s  —  a 

This  value  of  a  satisfies  the  two  special  cases  referred  to  ;  for 
in  the  first  case,  since  d  =  u  and  a  =  u,  we  must  have  a  —  i  ; 
and  if  we  write  a  =  u  in  the  value  of  a,  we  obtain  a  =  i.  Also 
in  the  second  case,  since  d  =  2s  when  a  =  s,  we  must  have 
a  =  i  ;  and  if  we  write  a  =  s  in  the  value  of  a,  we  obtain  a  =  |. 

This,  however,  has  not  been  tested  for  intermediate  values. 


WEYRAUCH'S  FORMULA.  505 

Now  write  d  =  a  -{-  c  in  equation  (i),  and  we  shall  have 

u  —  s 
a  =  -  (a  4-  c), 

2u   —   S   —   (T 

2au  —  as  —  a2  =  au  —  as  -f-  (u  —  s)c 

.-.     a2  —  au  —  (u  —  s)c  (7) 

•'•    «  =  «  -  («  -  f)j|  (8) 

and  if  we  write 

^  _  max  Z' 
a        max  Z' 

where  max  L  =  greatest  load  on  the  piece,  and  max  Lf  = 
greatest  load  of  opposite  kind,  so  that,  if  L  is  tension,  Lr  shall 
be  compression,  and  vice  versa,  we  shall  have 

v  max  Lr 

(9) 


this  being  in  a  form  suitable  to  use,  the  more  common  form 
being 


—  s  max  Lr  ) 
- 


i  ---    -  -\.  (10) 

u      max  L  ) 


The  values  of  the  constants  as  determined  from  Wohler's 
experiments,  and  the  resulting  form  of  the  formulae  for  Phoenix 
axle-iron  and  for  Krupp  cast-steel,  are  given  in  §  176. 

The  above  demonstrations  of  Launhardt's  and  of  Wey- 
rauch's  formulas  are  substantially  those  given  in  the  translation 
of  Weyrauch's  "  Structures  of  Iron  and  Steel,"  by  Professor 
Dubois,  the  explanations  being  somewhat  changed. 


506  APPLIED   MECHANICS. 


GENERAL    REMARKS. 

In  each  case  the  value  of  a  given  by  the  formula  (5)  or  (10) 
is  the  breaking-strength  per  unit  of  area  for  either  tension  or 
compression  in  Launhardt's  formula ;  and  in  the  case  of  Wey- 
rauch's,  the  value  of  a  is  the  breaking-strength  per  square  inch 
of  the  same  kind  as  max  L  ;  i.e.,  tension  if  this  is  tension,  or 
compression  if  this  is  compression. 

If  either  of  these  values  of  a  be  divided  by  3,  we  have, 
according  to  Weyrauch,  the  safe  working-strength. 

If,  now,  the  piece  be  a  tension  piece,  its  area  will  be  found 
by  dividing  its  greatest  load  by  the  safe  working-strength ;  if, 
on  the  other  hand,  it  be  subjected  to  compression,  and  it  be  a 
short  piece,  its  area  will  also  be  found  by  dividing  the  greatest 
compression  by  the  safe  working-strength  per  unit  of  area :  but 
if  it  be  a  long  column,  and  we  wish  to  use  Wohler's  results,  we 

must  merely  use  the  value  -  as  the  safe  working-strength  per 

unit  of  area,  and  use  this  in  whatever  formula  we  may  employ 
for  calculating  a  column. 

WOHLER'S   EXPERIMENTAL   RESULTS. 

Wohler  himself  made  his  tests  upon  the  extremes  of  fibre 
stresses  of  which  a  piece  could  bear,  without  breaking,  an 
unlimited  number  of  applications.  He  gives,  as  a  summary  of 
these  results,  the  following  :  — 

In  iron,  — 

Between  4-16000  Ibs.  per  sq.  in.  and  —16000  Ibs.  per  sq.  in. 
«       +30000       "  "        "  o  "  " 

+44000       "  "        "    +24000       "  " 

In  axle-steel,  — 

Between  +28000  Ibs.  per  sq.  in.  and  —28000  Ibs.  per  sq.  in. 

+  48000       "  "  o 

"       +80000       "  "        "    +35000       "          " 


WOHLER'S  EXPERIMENTAL   RESULTS.  507 

In  untempered  spring  steel,  — 

Between  +50000  Ibs.  per  sq.  in.  and          o      Ibs.  per  sq.  in. 

+  70000        "  "        "    +25000 

"        +80000       "  "        "     +40000        "  " 

+  90000        "  "        "    +60000       "  " 

For  shearing  in  axle-steel,  — 

Between  +22000  Ibs.  per  sq.  in.  and  —22000  Ibs.  per  sq.  in. 
+  38000        "  "        "  o 

This  table  would  justify  the  use,  in  Launhardt's  and  Wey- 
rauch's  formulae,  of  the  following  values  of  u  and  s  ;  viz., — 
In  iron,  — 

u  =  30000  Ibs.  per  sq.  in., 
s  =  16000  Ibs.  per  sq.  in. 

In  axle  steel,  — 

u  =  48000  Ibs.  per  sq.  in., 
j  =  28000  Ibs.  per  sq.  in. 

In  untempered  spring  steel,  — 

u  =  50000  Ibs.  per  sq.  in. 

And  it  would  require,  that  if,  with  these  values  of  u,  and  the 
values  of  t  given  in  §§  172  and  176,  we  put 

c  —  24000 

in   Launhardt's  formula  for  iron,  we  ought  to  obtain  approxi- 
mately 

a  =  44000 ; 

and  if  we  put  c  —  35000  in  that  for  steel,  we  should  obtain 
approximately 

a  =  80000. 


5O8  APPLIED   MECHANICS. 


FACTOR   OF   SAFETY. 

We  have  seen  that  Weyrauch  recommends,  to  use  with 
Wohler's  results,  a  factor  of  safety  of  three  for  ordinary  bridge 
work  and  similar  constructions. 

Wohler  himself,  however,  in  his  "  Festigkeits  versuche  mit 
Eisen  und  Stahl,"  says, — 

i°.  That  we  must  guard  against  any  danger  of  putting  on 
the  piece  a  load  greater  than  it  is  calculated  to  resist,  by  assum- 
ing as  its  greatest  stress  the  actually  greatest  load  that  can 
ever  come  upon  the  piece ;  and 

2°.  This  being  done,  that  the  only  thing  to  be  provided  for 
is  the  lack  of  homogeneity  in  the  material. 

3°.  That  any  material  which  requires  a  factor  of  safety 
greater  than  two  is  unfit  for  use.  This  advice  would  hardly  be 
accepted  by  engineers,  however. 

He  also  claims  that  the  reason  why  it  is  safe  to  load  car- 
springs  so  much  above  their  limit  of  elasticity,  and  so  near 
their  breaking-load,  is,  that  the  variation  of  stress  to  which  they 
are  subjected  is  very  inconsiderable  compared  with  the  greatest 
stress  to  which  they  are  subjected. 

GENERAL   REMARKS. 

It  is  to  be  observed,  — 

i°.  The  tests  were  all  made  on  a  good  quality  of  iron  and 
of  steel,  consequently  on  materials  that  have  a  good  degree  of 
homogeneity. 

2°.  The  specimens  were  all  small,  and  the  repetitions  of  load 
succeeded  each  other  very  rapidly,  no  time  being  given  for  the 
material  to  rest  between  them. 

3°.  No  observations  were  made  on  the  behavior  of  the  piece 
during  the  experiment  before  fracture. 

4°.  No  experiments  were  made  upon  cast-iron  and  timber ; 
and  the  results  of  such  experiments,  if  made,  could  hardly  be 


SHEARING-STRENGTH   OF  IRON  AND   STEEL.  509 

expected  to  be  of  much  value,  as  these  materials  are  so  lacking 
in  homogeneity. 

5°.  As  long  as  we  are  dealing  only  with  tension,  we  can  say 
without  error  that 

c_  •_  min  Z  . 

a       max  L ' 

but  as  soon  as  both  stresses  or  either  become  compression,  if 
the  piece  is  long  compared  with  its  diameter,  we  cannot  assert 
with  accuracy  the  above  relation,  nor  that 

c  _  max  U  t 
a       maxZ  ' 

and  hence  results  based  on  these  assumptions  must  be  to  a 
certain  extent  erroneous. 

6°.  When  a  piece  is  subjected  to  alternate  tension  and  com- 
pression, it  must  be  calculated  so  as  to  bear  either :  thus,  if 
sufficient  area  is  given  it  to  enable  it  to  bear  the  tension,  it  may 
not  be  able  to  bear  the  compression  unless  the  metal  is  so  dis- 
tributed as  to  enable  it  to  withstand  the  bending  that  results 
from  its  action  as  a  column. 

EXAMPLES. 

1.  Determine   the   cross-section   necessary  for   a   wrought-iron   tie 
where  greatest  load  =  800000  Ibs.  and  least  load  =  80000  Ibs. 

2.  Determine  the  cross-section  necessary  for  a  wrought-iron  strut  to 
bear  the  same  loads. 

3.  What  is  the  greatest  and  what  the  least  working-strength  recom- 
mended by  Wohler  for  wrought-iron  ?     What  for  steel  ?     Compare  with 
ordinary  methods,  using  factor  of  safety  of  four. 

BAUSCHINGER'S  TESTS  ON  REPEATED  STRESSES. 

Bauschinger's  definitions  of  elastic  limit  and  of  stretch 
limit  have  already  been  given.  He  also  states  that  other 
properties  of  the  metal  at  or  near  the  elastic  limit  are  the  fol- 
lowing, viz.  : 


510  APPLIED   MECHANICS. 


(a)  The  sets  within  the  elastic  limit  are  very  small,  and  in- 
crease proportionally  to  the  load,  while  above  that  point  they 
increase  much  more  rapidly. 

(b)  With  repeated  loading,  inside  of  the  elastic  limit,  drop- 
ping to  zero  between  times,  we  find  each  time   the  same  total 
elongations. 

(c)  While  within  the   elastic  limit  the  elongations  remain 
constant  as  long  as  the  load  is  constant ;  with  a  load  above  the 
elastic   limit   the   final   elongations   under  that  load  are   only 
reached  after  a  considerable  length  of  time. 

(d)  He  also  says  that  substances  like  cast-iron  and  stone,  in 
which  the  stress  is  nowhere  proportional  to  the  strain,  have  no 
elastic  limit, 

Bauschinger  gives  as  consequences  of  his  definition  of  elas- 
tic limit  the  following  : 

If  by  subjecting  a  rod  to  changing  stresses  between  an  upper 
and  lower  limit,  of  which  at  least  the  upper  is  above  the  original 
elastic  limit,  the  latter  were  either  unchanged  or  lowered,  or  if 
in  the  case  of  its  being  raised  it  were  to  remain  below  the  upper 
limit,  then  the  repetition  of  such  stresses  must  finally  end  in 
rupture,  for  each  new  application  of  the  stress  increases  the 
strain  ;  but  if  both  limits  of  the  changing  stress  are  and  remain 
below  the  elastic  limit,  the  repetition  will  not  cause  break- 
age. 

Bauschinger  says  that  by  overstraining,  the  stretch  limit  is 
always  raised  up  to  the  load  with  which  the  stretching  was 
done ;  but  in  the  time  of  rest  following  the  unloading  the 
stretch  limit  rises  farther,  so  that  it  becomes  greater  than  the 
maximum  load  with  which  the  piece  was  stretched,  and  this 
rising  continues  for  days,  months,  and  years  ;  but,  on  the  other 
hand,  that  the  elastic  limit  (according  to  his  definition)  is  low- 
ered by  the  overstraining,  often  to  zero  ;  and  that  a  subsequent 
rest  gradually  raises  it  until  it  reaches,  after  several  days,  the 
load  applied,  and  in  time  rises  above  this  ;  that,  as  a  rule,  the 


EXPERIMENTS  WITH  A  REPEATED   TENSION  MACHINE     $1 1 

modulus  of  elasticity  is  also  lowered  under  the  same  circum- 
stances, and  is  also  restored  by  rest,  and  rises  after  several 
years  above  its  original  magnitude. 

The  following  conclusions,  drawn  by  Bauschinger  from  his 
experiments,  are  of  very  great  importance  : 

i°.  By  a  tensile  load  above  the  elastic  limit  the  elastic  limit 
for  compression  is  lowered,  and  vice  versa  for  a  compressive 
load  ;  and  a  comparatively  small  excess  over  the  elastic  limit 
for  one  kind  of  load  may  lower  that  for  the  opposite  kind 
down  to  zero  at  once.  Moreover,  an  elastic  limit  which  has 
been  lowered  in  this  way  is  not  materially  restored  by  a  period 
of  rest — at  any  rate,  of  three  or  four  days. 

2°.  With  gradually  increasing  stresses,  changing  from  ten- 
sion to  compression,  and  vice  versa,  the  first  lowering  of  the 
elastic  limit  occurs  when  the  stresses  exceed  the  original  elastic 
limit. 

3°.  If  the  elastic  limit  for  tension  or  compression  has  been 
lowered  by  an  excessive  load  of  the  opposite  kind,  i.e.,  one  ex- 
ceeding the  original  elastic  limit,  then,  by  gradually  increasing 
stresses,  changing  between  tension  and  compression,  it  can 
again  be  raised,  but  only  up  to  a  limit  which  lies  considerably 
below  the  original  elastic  limit. 

EXPERIMENTS   WITH   A  REPEATED   TENSION,  MACHINE. 

Bauschinger  states  that  in  1881  he  acquired  a  machine 
similar  to  that  used  by  Wohler  for  repeated  application  of  a 
tensile  stress. 

The  plan  of  the  experiments  which  he  made  with  it,  and 
which  are  detailed  in  the  I3th  Heft  of  the  Mittheilungen,  is  as 
follows : 

From  a  large  piece  of  the  material  there  were  cut  at  least 
four,  and  sometimes  more,  test-pieces  for  the  Wohler  machine. 
One  of  them  was  tested  in  the  Werder  machine  to  determine 
its  limit  of  elasticity  and  its  tensile  strength  ;  the  others  were 


512  APPLIED  MECHANICS. 

tested  in  the  Wohler  machine,  so  arranged  that  the  upper  limit 
of  the  repeated  stress  should  be,  for  the  first  specimen,  near 
the  elastic  limit  ;  for  the  second,  somewhat  higher,  etc.,  the 
lower  limit  being  in  all  cases  zero. 

From  time  to  time  the  test-pieces,  after  they  had  been  sub- 
jected to  some  hundred  thousands,  or  some  millions,  of  repeti- 
tions, were  taken  from  the  Wohler  machine  and  had  their  limits 
of  elasticity  determined  in  the  Werder  machine. 

The  following  were  the  kinds  of  specimens  tested  : 

i°.  Six  rectangular  bars  from  a  wrought-iron  plate  n  mil- 
limeters thick. 

2°.  Seventeen  rectangular  bars  from  a  mild  steel-plate  n 
mm.  thick. 

3°.  Six  rectangular  bars  cut  from  an  So-mm.  by  lo-mm.  bar. 

4°.  Four  rectangular  bars  from  a  4O-mm.  by  lo-mm.  bar. 

5°.  Four  round  specimens  from  a  broken  piece  of  an  axle 
of  Thomas  steel  12.8  cm.  thick,  which  had  been  tested  for  bend- 
ing-strength. 

6°.  Four  round  specimens  from  the  heads  of  the  broken 
pieces  of  a  rail  of  Thomas  steel,  which  had  been  tested  for 
transverse  strength. 

7°.  Eight  rectangular  bars  from  a  12-mm.  thick  boiler-plate 
of  mild  Thomas  steel. 

The  tables  of  the  tests  are  to  be  found  in  the  Mittheilungen, 
and  from  them  Bauschinger  draws  the  following  conclusions : 

1°.  With  repeated  tensile  stresses,  whose  lower  limit  was 
zero,  and  whose  upper  limit  was  near  the  original  elastic  limit, 
breakage  did  not  occur  with  from  5  to  16  millions  of  repeti- 
tions. 

Bauschinger  says  that  in  applying  this  law  to  practical  cases 
we  must  bear  in  mind  two  things  :  (a)  that  it  does  not  apply 
when  there  are  flaws,  as  several  specimens  which  contained  flaws, 
many  of  them  so  small  as  to  be  hardly  discoverable,  broke  with 
a  much  smaller  number  of  repetitions;  (#)  another  caution  is 


EXPERIMENTS  WITH  A  REPEATED   TENSION  MACHINE.   513 

that  we  should  make  sure  that  we  know  what  is  really  the  origi- 
nal elastic  limit,  as  this  varies  very  much  with  the  previous. 
treatment  of  the  piece,  especially  the  treatment  it  received 
during  its  manufacture,  and  it  may  be  very  small,  or  it  may  be 
very  near  the  breaking-strength. 

2°.  With  oft-repeated  stresses,  varying  between  zero  and  art 
upper  stress,  which  is  in  the  neighborhood  of  or  above  the 
original  elastic  limit,  the  latter  is  raised  even  above,  often  far 
above,  the  upper  limit  of  stresses,  and  the  higher  the  greater 
the  number  of  repetitions,  without,  however,  its  being  able  to 
exceed  a  known  limiting  value. 

3°.  Repeated  stresses  between  zero  and  an  upper  limit, 
which  is  below  the  limiting  value  of  stress  which  it  is  possible 
for  the  elastic  limit  to  reach,  do  not  cause  rupture  ;  but  if  the 
upper  limit  lies  above  this  limiting  value,  breakage  must  occur 
after  a  limited  number  of  repetitions. 

4°.  The  tensile  strength  is  not  diminished  with  a  million 
repetitions,  but  rather  increased,  when  the  test-piece  after  hav- 
ing been  subjected  to  repeated  stresses  is  broken  with  a  steady 
load. 

5°.  He  discusses  here  the  probability  of  the  time  of  forma- 
tion of  what  he  considers  to  be  a  change  in  the  structure  of 
the  metal  at  the  place  of  the  fracture. 

Besides  the  above  will  be  given  the  numerical  values  which 
Bauschinger  obtained  for  carrying  strength  and  for  primitive 
safe  strength  as  average  values. 

i°.  For  wrought-iron  plates: 

/  —  49500  Ibs.  per  sq.  in. 

u  =  28450    " 


"     "     " 


2°.  For  mild-steel  plates  (Bessemer)  : 
/  —  62010  Ibs.  per  sq.  in. 


u  =  34140 


"      "     "     " 


5  14  APPLIED   MECHANICS. 

3°.  For  bar  wrought-iron,  80  mm.  by  10  mm. : 

t  =  57600  Ibs.  per  sq.  in. 
u—  31290    "      "     "     " 

4°.  For  bar  wrought-iron,  40  mm.  by  10  mm. : 

t  —  57180  Ibs.  per  sq.  in. 
u  =  34140    "      "     "     " 

5°.  For  Thomas-steel  axle  : 

/  =  87050  Ibs.  per  sq.  in. 
u  —  42670    "      "     "     " 

6°.  For  Thomas-steel  rails  : 

/  =  84490  Ibs.  per  sq.  in. 
u  —  39820    "      "     "     " 

7°.  For  Thomas-steel  boiler-plate  : 

t  —  57600  Ibs.  per  sq.  in. 
u  =  34140    "      "     "     " 

§  229.  Shearing-Strength  of  Iron  and  Steel.— Some  of 
the  most  common  cases  where  the  shearing  resistance  of  iron 
and  steel  is  brought  into  play  are  :  — 

i°.  In  the  case  of  a  torsional  stress,  as  in  shafting. 

2°.  In  the  case  of  pins,  as  in  bridge-pins,  crank-pins,  etc. 

3°.  In  the  case  of  riveted  joints. 

In  most  cases  the  shearing  is  accompanied  by  tensions  or 
compressions,  or  other  stresses,  and  it  is  difficult  to  separate 
the  effects ;  so  that,  in  the  present  state  of  our  knowledge,  we 
cannot  explain  the  fact,  that,  when  the  breaking  shearing- 
strength  of  wrought-iron  or  steel  is  deduced  from  an  experiment 
on  torsional  strength  it  is  found  to  be  about  equal  to  the  ten- 


TOR  SIGNAL  STRENGTH  OF  WROUGHT-IRON,  ETC.       5 15 

sile  strength,  while  when  deduced  from  experiments  on  riveted 
joints,  it  is  found  to  be  about  f  or  |-  the  tensile  strength. 

In  regard  to  cast-iron,  Bindon  Stoney  found  the  shearing 
and  tensile  strength  about  equal. 

As  to  shearing  modulus  of  elasticity,  Bauschinger's  experi- 
ments on  cast-iron  give  about  two-fifths  the  tensile  modulus  of 
elasticity ;  and  Wohler's  experiments  gave  for  steel  almost 
exactly  two-fifths  the  tensile  modulus,  his  value  being  11236500 
when  the  tensile  modulus  was  29600000.  According  to  the 
theory  of  elasticity,  the  modulus  of  elasticity  for  shearing 
should  be  about  two-fifths  that  for  tension  or  compression,  as 
will  be  shown  later;  and  these  experiments  furnish  a  most 
beautiful  confirmation  of  the  theory  of  elasticity. 

The  cases  where  shearing  comes  in  play  in  wrought-iron  and 
steel  will,  therefore,  be  treated  separately. 

§  230.  Torsional  Strength  of  Wrought-iron  and  Steel.  — 
The  most  common  custom  for  computing  the  strength  of  a 
shaft  has  been  to  use  one  based  upon  its  twisting-moment ;  and 
hence  using  the  shearing  breaking-strength  of  the  material,  as 
determined  from  an  experiment  on  simple  torsion,  for  our  con- 
stant. It  is  generally  the  fact,  however,  that,  when  shafting  is 
running,  the  pulls  of  the  belts  create  a  bending  backwards  and 
forwards,  bringing  the  same  fibre  alternately  into  tension  and 
compression ;  and  this  is  combined  with  the  shearing-stresses 
developed  due  to  the  twisting-moment  alone.  At  the  two  ex- 
tremes of  these  general  cases  are  :  — 

i°.  The  case  when  the  portion  of  a  shaft  between  two 
hangers  has  no  pulleys  upon  it,  and  when  the  pulls  on  the 
neighboring  spans  are  not  so  great  as  to  deflect  this  span  appre 
ciably.  That  is  a  case  of  pure  torsion  :  and  if  the  shaft  is  run* 
ning  smoothly,  with  no  jars  or  shocks,  and  no  liability  to  have 
a  greater  load  thrown  upon  it  temporarily,  we  may  compute  it 
by  the  usual  torsion  formula,  given  in  §  212  ;  using  for  break- 
ing-strength of  wrought-iron  and  steel  the  tensile  strength,  and 


APPLIED  MECHANICS. 


a  factor  of  safety  six,  and  such  a  proceeding  will  probably  give 
us  a  reasonable  degree  of  safety. 

2°.  The  case  when,  pulleys  being  placed  otherwise  than  near 
the  hangers,  the  belt-pulls  are  so  great  that  the  torsion  becomes 
insignificant  compared  with  the  bending,  and  then  it  would  be 
proper  to  compute  our  shaft  so  as  not  to  deflect  more  than  y^Vir 
of  its  span  under  the  load,  or  better,  not  more  than  y^Vir  :  of 
course  we  should  compute  also  the  breaking  transverse  load, 
and  see  that  we  have  a  good  margin  of  safety. 

In  other  cases,  the  methods  pursued  up  to  the  present  time 
have  been  the  following  ;  viz.  :  — 

i°.  By  using  the  ordinary  torsion  formula  combined  with  a 
large  factor  of  safety. 

2°.  As  recommended  by  D.  K.  Clark,  by  computing  the 
shaft  also  for  deflection,  and  providing  that  its  deflection  shall 
not  exceed  y^  or  y^rr  of  its  span. 

This,  however,  neglects  the  torsion,  and  also  the  rapid 
change  of  stress  upon  each  fibre  from  tension  to  compression, 
and  vice  versa. 

3°.  By  using  the  formula  of  Grashof  or  of  Rankine  for  com- 
bined bending  and  twisting,  with  the  constants  that  have  been 
derived  from  experiments  on  simple  tension  or  simple  torsion. 

In  regard  to  experiments  upon  torsional  strength,  they  are 
usually  made  by  twisting  a  short  piece  of  shaft  until  rupture 
occurs,  measuring  its  twist  under  smaller  loads,  and  from  these 
computing  the  modulus  of  shearing  elasticity. 

Such  tests  have  been  made  by  — 

i°.  Kirkaldy  :  D.  K.  Clark's  Rules  and  Tables. 
2°.  D.  K.  Clark  :  English  Civil  Engineers'  Committee. 
3°.  Major  Wade. 

4°.  United-States  Board  to  test  Iron  and  Steel:  Executive  Docu- 
ment No.  23,  46th  Congress,  2d  session. 

5°.  Professor  Thurston  :  Materials  of  Engineering. 


TOR  SIGNAL  STRENGTH  OF  WROUGHT-IRON,  ETC.       5  1/ 

The  general  result  has  usually  been,  to  obtain  about  the 
same  shearing  breaking  fibre  stress  as  the  tensile  strength  of 
the  material,  and  a  modulus  of  elasticity  about  two-fifths  the 
tensile  modulus.  Of  course  the  values  vary  more  or  less,  some- 
times being  greater  and  sometimes  less  than  the  values  given 
above,  according  to  the  quality  of  the  iron  ;  so  that  we  can  de- 
termine by  experiment  only  what  any  one  iron  will  bear.  None 
of  these  tests  will  be  recorded  here,  but  some  of  the  rules  in 
common  use  for  figuring  the  strength  will  be  given  :  they  are 
merely  some  of  the  formulae  already  referred  to,  with  the  con- 
stants changed. 

Under  certain  circumstances  the  bending  may  have  the 
greatest  influence,  while  the  twisting  may  be  predominant  in 
others,  or  their  influence  may  be  equally  divided.  Which  of 
these  is  the  case  will  depend  upon  the  location  of  the  hangers 
and  of  the  pulleys,  the  width  of  the  belts,  etc.,  etc. 

As  to  the  formulae  which  take  into  account  both  twisting 
and  bending,  there  are  two,  both  of  which  are  based  upon  the 
theory  of  elasticity.  The  first,  which  is  the  most  correct  from 
a  theoretical  point  of  view,  is  that  given  by  Grashof  and  other 
writers  on  the  theory  of  elasticity,  and  is 


where  Ml  =  greatest  bending  moment  ; 
MI  =  greatest  twisting  moment  ; 
r  =  external  radius  of  shaft  ; 

/=  moment  of  inertia  of  section  about  a  diameter; 
f  =  greatest  allowable  stress  at  outside  fibre  ; 
m  =  a  constant  depending  on  the  nature  of  the  ma- 
terial. 

If,  as  is  usually  done  for  iron  and  steel,  we  make  m  =  4,  we 
obtain 


Sl8  APPLIED  MECHANICS. 


The  other  formula,  which  is  also  based  upon  the  theory  of 
elasticity,  but  which  is  not  as  correct,  is  that  given  by  Rankine, 
and  is 


While  these  formulae  are  the  only  ones  that  take  both 
twisting  and  bending  into  account,  they  are  not  very  generally 
used,  being  recommended  by  some  engineers  for  use  in  crank- 
shafts and  propeller-shafts  of  marine  engines,  while  they  are 
seldom  used  elsewhere. 

The  formulae  in  general  use  for  line-shafting  and  head- 
shafting  are  most  commonly  those  taking  into  account  the 
twisting  only,  and  the  constants  for  these  formulae  are  those 
deduced  from  experiments  upon  pure  torsion.  Of  this  nature 
are  the  following  formulae  given  by  Prof.  Thurston : 

For  head-shafts  well  supported  against  springing, 

3 /I2-  jj  p  3 

Wrought-iron,  d  =  y  — **      '      ;     Cold-rolled  iron,  d  = 

For  line-shafting,  hangers  8  feet  apart, 
Wrought-iron,  d  =  \  - —    '      ;      Cold-rolled  iron,  d  = 

For  transmission  simply,  no  pulleys, 


Wrought-iron,  d  =  ''  ;    Cold  rolled  iron,  d  = 

Unwin  gives  the  direction,  — 

i°.  That  the  axle  must  be  calculated  as  a  beam  to  bear  the 
weight  of  the  pulley  and  the  belt-pull  that  is  to  come  upon  it. 


TORSIONAL   STRENGTH  OF   WROUGHT- 1  RON,   ETC.        519 

2°.  For  shafting  transmitting  power,  and  subject  to  torsion 
only,  he  gives 

'HP 


where  d  =  diameter  in  inches,  PR  =  twisting-moment  in  inch- 
pounds,  N  =  number  of  turns  per  minute,  HP  =  number  of 
horses-power  to  be  transmitted;  and  he  gives  for  a  and  ft  the 
following  values :  — 


a. 

£ 

0.0827^ 

•2.2Q4 

0.1042 

4.  ICQ 

Steel  

0.072^ 

•r*<O** 

2.877 

these  values  assuming  for  safe  greatest  fibre  stress,  for  wrought- 
iron,  9000;  for  cast-iron,  4500;  and  for  steel,  13500  Ibs.  per 
square  inch. 

3°.  For  the  crank-shaft  of  a  steam-engine  he  advises  us  to 
consider  the  maximum  twisting-moment  as  1.3  times  the  mean. 

4°.  For  combined  twisting  and  bending,  he  gives  the  Ran- 
kine  formula,  equation  (2),  §  214,  and  puts  it  in  the  following 
form : 


where  d  =.  diameter  in  inches,  f  =  outside  fibre  breaking  shear- 
ing-strength per  square  inch,   T  =  twisting-moment  in  inch- 

M 
pounds,  K  =  —  =  ratio  of  bending  to  twisting  moment. 

Then,  calling 


$20  APPLIED   MECHANICS. 

n  will  be  the  ratio  of  the  diameter  to  be  used,  to  the  diameter 
needed  to  resist  the  torsion  alone ;  and  he  gives,  when 

K  =  0.25     0.50     0.75     i.oo     1.25     1.50     1.75     2.00     3.00, 
n  =  1.09     1.17     1.26     1.36     1.42     1.49     1.56     1.62     1.83. 

For  the  propeller  shaft  of  a  steam-vessel,  he  advises  to  use 

K  =  0.25  to  0.50  /.     n  =  1.09  to  1.17. 

For  line  shafting  in  mills,  he  advises 

K  =  0.75  to  i.oo  /.     n  =  1.26  to  1.34. 

For  crank-shafts  and  heavy  shafting  subject  to  shocks, 

K  =  i.oo  to  1.50  /.     n  =  1.34  to  1.49. 

5°.  For  forged  crank-shafts,  he  gives 

d  =  4-55 

To  control  the  deflection  of  shafting  10  or  n  feet  between 
hangers,  he  gives 


where  L  ='span  in  inches,  and  y  =  54  to  60. 

With  a  view  to  determine  the  behavior  of  shafting  under  a 
combination  of  twisting  and  bending,  suitable  machinery  was 
erected  in  the  engineering  laboratories  of  the  Mass.  Institute 
of  Technology,  and  a  number  of  tests  were  made. 

The  principal  points  of  the  method  of  procedure  are  the 
following,  viz. : 

1st.  The  shaft  under  test  is  in  motion,  and  is  actually  driv- 
ing an  amount  of  power  which  is  weighed  on  a  Prony  brake. 

2d.  A  transverse  load  is  applied  which  may  be  varied  at 
the  option  of  the  experimenter,  and  which  is  weighed  on  a 
platform  scale. 


TORSIONAL    STRENGTH  OF    WROUGHT-IRON,   ETC.      $21 


3d.  The  proportion  between  the  torsional  and  transverse 
loads  may  be  adjusted  to  correspond  with  the  proportion  be- 
tween the  power  transmitted  and  the  belt-pull  sustained  by  a 
shaft  in  actual  use. 

4th.  Tests  are  made  not  only  of  breaking-strength,  but  also 
angle  of  twist  and  deflection  under  moderate  loads  are  measured. 

The  following  table  will  give  the  results  of  the  tests  on 
iron  shafts,  and  they  will  then  be  discussed  : 


Time 

'*•„ 

4f, 

/,, 

/», 

No. 

of 

Total 

H.  P. 

max. 

max. 

max. 

max. 

j- 

of 
Test. 

run- 
ning, 
min- 

revolu- 
tions. 

trans- 
mitted. 

bending 
moment. 

twisting 
moment. 

bend, 
fibre 

twist, 
fibre 

Grashof. 

Ran- 

kine. 

Diam. 
ins. 

utes. 

In.-lbs. 

In.-lbs. 

stress. 

stress. 

8 

37-5 

7040 

11.717 

11514.1 

3926.4 

60024 

10234 

62162 

61755 

l".25 

9 

200 

38839 

8.181 

10507.8 

2656.8 

54777 

6925 

55876 

55671 

".25 

10 

162 

31641 

5-291 

9891.0 

1714.6 

51562 

4469 

52062 

51976 

".25 

ii 

553 

.108002 

4-331 

9241.7 

1399.2 

48179 

3647 

48539 

48769 

".*5 

12 

V8 

80694 

6.276 

9241.7 

2027.6 

48179 

5287 

48911 

48769 

"•25 

13 

98 

19333 

6.342 

8917.1 

2028  .  2 

46485 

5287 

47245 

47105 

"•25 

14 

423 

82741 

6.283 

8917.1 

2029.7 

46485 

5290 

47246 

47106 

".25 

15 

565 

108739 

6.192 

8592-5 

2031.6 

44793 

5295 

45582 

45436 

".25 

16 

443 

88208 

6.338 

8267.8 

2026.8 

43100 

5283 

439M 

43713 

".25 

17 

95i 

185233 

6.283 

3781.5 

2029.7 

38503 

"333 

41768 

41117 

I" 

14.834 

8218 

84185 

i 

68 

It 

20 

7  .562 

7076 

82112 

12188 

8  81 

8  o  i 

It 

21 

9O72 

/y/v' 
8017 

4  19 

3031 

n 

22 

•yr* 

oyi/ 
8oi7 

4848 

Q 

10454 

94434 
08612 

9371" 

n 

' 

°y*/ 

90793 

24     i 

it 

23 

2-955 

7  52 

945 

779I3 

4  " 

783M 

70239 

In  19  to  23  inclusive  the  number  of  revolutions  was  small  and 
the  outside  fibre  stress  at  fracture  was  correspondingly  large. 

Two  specimens  of  the  i".25  shafting  and  two  of  the  i" 
were  tested  for  tension,  the  results  being  as  follows : 

Breaking-strength,  per  sq.  in. 

diameter      N-    ;;;;;;     468oo 


Average 


•    48333 


APPLIED  MECHANICS. 


Average     ....      60250 
As  to  conclusions  : 

1st.  It  is  plain  from  these  results  that  a  shaft  whose  size  is 
determined  by  means  of  the  r<  suits  of  a  quick  test  would  be 
too  weak,  and  that  our  constants  should  be  obtained  from  tests 
which  last  for  a  considerable  length  of  time. 

2d.  A  perusal  of  the  tables  will  show  that  the  results  ob- 
tained apply  more  to  the  bending  than  to  the  twisting  of  a 
shaft,  as  the  transverse  load  used  in  these  tests  was  so  large 
compared  with  the  twist  as  to  exert  the  controlling  influence. 
This  will  be  plain  by  a  comparison  of  the  values  of  /,  ,/3, 
and/". 

3d.  Nevertheless,  the  bending-moments  actually  used  were 
generally  less  than  such  as  might  easily  be  realized  in  practice 
with  the  twisting-moments  used.  * 

4th.  It  seems  fair  to  conclude  that,  in  the  greater  part  of 
cases  where  shafting  is  used  to  transmit  power,  as  in  line-shaft- 
ing or  in  most  cases  of  head  shafting,  the  breaking  is  even  more 
liable  to  occur  from  bending  back  and  forth  than  from  twist- 
ing, and  hence  that  in  no  such  case  ought  we  to  omit  to 
make  a  computation  for  the  bending  of  the  shaft  as  well  as  the 
twist. 

5th.  As  to  the  precise  value  of  the  greatest  allowable  out- 
side fibre  stress  to  be  used  in  the  Grashof  formula,  it  is  plain 
that  it  is  not  correct  to  use  a  value  as  great  as  the  tensile 
strength  of  the  iron,  and  while  the  tests  show  that  this  figure 
should  not  for  common  iron  exceed  40000  Ibs.  per  square  inch, 
it  is  probable  that  tests  where  a  longer  time  is  allowed  for 
fracture  will  show  a  smaller  result  yet. 

TWIST   UNDER   MODERATE  LOADS. 

Next,  as  to  twist  under  moderate  loads,  the  results  give  us 


TWIST   UNDER  MODERATE  LOADS. 


523 


the  modulus  of  elasticity  for  torsion.     The  values  were  ob- 
tained from  experiments  on  i"  shafts,  and  are  as  follows:  — 


No.  of  Test. 

^No.  of  Test. 

18  . 

11268588 

24   .   . 

.   11293188 

19  . 

9979319 

25   •   • 

.   12878875 

20  . 

15911690 

26   .   . 

.   13410870 

21   . 

22   . 

23   • 

12169442 
I222I239 
11943088 

9)111076299 
12341811 

The  tensile  modulus  of  elasticity  of  this  shafting,  as  deter- 
mined by  experiment,  was  29008621  pounds  per  square  inch. 
^  The  theory  of  elasticity  gives  the  shearing  modulus  of  elas- 
ticity as  two-fifths  the  tensile,  or  f  (29008621)=  11603448, 
which  is  not  far  from  the  average  value  found  by  experiment. 

Mr.  James  B.  Francis  gives  the  following  table  for  distance 
between  bearings  of  shafting  carrying  no  side  strain  : — 


Diameter. 

Wrought-Iron. 

Steel. 

Diameter. 

Wrought-Iron. 

Steel. 

in. 

ft. 

ft. 

in. 

ft. 

ft. 

2 

IS-S 

15-9 

6 

22.3 

22.9 

3 

17.7 

18.2 

7 

23-5 

24.1 

4 

19-5 

2O.O 

8 

24.6 

25.2 

5 

20.9 

21.6 

9 

25-5 

26.2 

It  should  be  observed,  that  it  is  only  a  mild  steel  that  can 
be  used  for  shafting,  —  steel  of  about  60000  Ibs.  tensile  strength 
per  square  inch :  hence  the  calculations  cannot  differ  very 
greatly  from  those  for  wrought-iron. 

§  231.  Bridge-Pins.  —  In  the  case  of  a  bridge-pin  or  other 
pin,  the  shearing-stress  is  always  accompanied  by  a  very  large 
element  of  bending ;  so  that  the  principles  of  transverse 
strength  come  into  play  very  largely,  and  perhaps  in  many 
cases  wholly. 


524  APPLIED  MECHANICS. 

The  rules  for  proportioning  bridge-pins  used  by  different  en- 
gineers and  constructors  have  been  various,  but  have  generally 
been  founded  upon  the  theory  of  transverse  strength  ;  for  those 
given  by  Charles  Bender,  the  student  is  referred  to  his  treatise 
•on  "Bridge-Pins"  (Van  Nostrand's  Science  Series).  No  attempt 
will  be  made  to  give  any  of  the  rules  used ;  but,  inasmuch  as 
there  were  quite  a  number  of  such  pins  tested  at  the  Watertown 
Arsenal,  there  will  be  given  a  summary  of  the  tests,  which  are 
to  be  found  in  detail  in  Executive  Document  12,  4/th  Congress, 
first  session.  All  the  pins  tested  were  of  wrought-iron,  and 
had  semicircular  seats  at  middle  and  ends. 

From  this  table  the  modulus  of  elasticity  can  be  figured, 
and  the  deflection  of  the  pin  under  any  given  load  and  span ; 
or  the  table  may  be  employed  to  interpolate,  to  find  the  sizes 
-to  be  used  without  causing  too  much  deflection. 


BRIDGE-PINS. 


52$ 


Diameter, 
in  inches. 

Clear 
Span. 

Width 
of 
Middle 
Bear- 
ing. 

Load  nearest 
one-half 
Maximum 
Load,  in  Ibs. 

Deflection, 
in  inches. 

Maximum 
Load  ap- 
plied, in  Ibs. 

Deflection, 
in  inches. 

Amount  of 
Indentation, 
in  inches. 

2.500 

24 

I 

IOOOO 

0.0517 

20000 

I.27OO 

0.0070 

2.497 

24 

I 

8000 

0.0420 

15000 

0.3290 

O.OO2O 

2.500 

18 

I 

IOOOO 

0.0238 

21000 

O.I  660 

O.OO25 

2.5*0 

18 

I 

IOOOO 

0.0248 

18000 

0.1000 

<O.OOIO 

2.500 

12 

- 

I5OOO 

0.0133 

3OOOO 

0.0367 

0 

2.500 

12 

- 

14000 

0.0129 

27000 

0.0437 

<  O.OOIO 

2.500 

6 

I 

33000 

0.0066 

65000 

0.0263 

0.0030 

2.500 

6 

I 

30000 

0.0064 

60000 

0.0184 

- 

3.000 

24 

- 

IIOOO 

0.0278 

22OOO 

0.1260 

- 

3-000 

24 

- 

IIOOO 

0.0278 

22000 

0.1231 

- 

3.OOO 

12 

If 

25000 

O.OIOI 

50000 

0.0414 

- 

3.OOO 

12 

If 

25000 

O.OIO2 

50000 

0.0464 

- 

3-OOO 

6 

- 

62000 

0.0077 

125000 

0.0240 

- 

3-000 

6 

If 

60000 

0.0070 

125000 

0.0250 

- 

3495 

24 

If 

2OOOO 

0.0302 

40000 

0.1681 

- 

3495 

24 

If 

2OOOO 

0.0310 

35000 

0.0923 

- 

3495 

12 

- 

40000 

O.OI2I 

80000 

0.0330 

- 

3495 

12 

- 

4OOOO 

O.OI24 

80000 

0.0340 

- 

3496 

6 

- 

90000 

0.0083 

I7OOOO 

0.0222 

O.OO4O 

3496 

6 

- 

9OOOO 

0.0087 

I6OOOO 

O.OI7I 

- 

4.000 

24 

If 

25000 

O.O227 

50000 

0.0667 

- 

4.000 

>  24 

If 

25OOO 

0.0234 

50000 

0.0682 

- 

4.000 

12 

4 

55000 

0.0097 

IIOOOO 

0.0258 

- 

4.000 

12 

if 

55000 

O.OI2O 

I  IOOOO 

0.0255 

- 

4.500 

24 

it 

38000 

O.O24O 

75000 

0.0648 

- 

4.500 

24 

if 

40000 

O.O26O 

75000 

0.0630 

- 

4.500 

12 

- 

80000 

0.0099 

160000 

0.0245 

- 

4.500 

12 

- 

80000 

0.0096 

170000 

0.0281 

- 

4.877 

24 

if 

50000 

O.O24O 

IOOOOO 

O.O9I2 

- 

4.877 

24 

if 

4OOOO 

O.OIOX) 

85000 

0.0460 

- 

4.877 

12 

if 

lOOOOO 

0.0155 

200000 

0.0324 

- 

4.877 

12 

if 

I  00000 

0.0152 

200000 

O.O322 

— 

526  APPLIED   MECHANICS. 


§  232.  Riveted  Joints.  —  The  most  common  way  of  uniting 
plates  of  wrought-iron  or  steel  is  by  means  of  rivets.  It  is, 
therefore,  a  matter  of  importance  to  know  the  strength  of  such 
joints,  and  also  the  proportions  which  will  render  their  efficien- 
cies greatest ;  i.e.,  that  will  bring  their  strength  as  near  as 
possible  to  the  strength  of  the  solid  plate. 

In  §  177  was  explained  the  mode  of  proportioning  riveted 
joints  usually  taught,  based  upon  the  principle  of  making  all 
the  resistances  to  giving  way  equal,  and  assuming,  as  the  modes 
of  giving  way,  those  there  enumerated.  This  theory  does  not, 
however,  represent  the  facts  of  the  case,  as  — 

i°.  The  stresses  which  resist  the  giving- way  are  of  a  more 
complex  nature  than  those  there  assumed,  so  that  the  efficiency 
of  a  joint  constructed  in  the  way  described  above  may  not  be 
as  great  as  that  of  one  differently  constructed ; 

2°.  The  effects  of  punching,  drilling,  and  riveting,  come  in 
to  modify  further  the  action  ;  and 

3°.  The  purposes  for  which  the  joint  is  to  be  used,  often  fix 
some  of  the  dimensions  within  narrow  limits  beforehand. 

In  order  to  know,  therefore,  the  efficiency  of  any  one  kind 
of  joint,  we  must  have  recourse  to  experiment.  And  here  again 
we  must  not  expect  to  draw  correct  conclusions  from  experi- 
ments made  upon  narrow  strips  of  plate  riveted  together  with 
one  or  two  rivets  ;  but  we  need  experiments  upon  joints  in  wide 
plates  containing  a  sufficiently  long  line  of  rivets  to  bring  into 
play  all  the  forces  that  we  have  in  the  actual  joint.  The  greater 
part  of  the  experiments  thus  far  made  have  been  made  upon 
narrow  strips,  with  but  few  rivets.  The  number  of  tests  of  the 
other  class  is  not  large,  and  of  those  that  have  been  made,  the 
greater  part  merely  furnish  us  information  as  to  the  behavior 
of  the  particular  form  of  joint  tested,  and  do  not  teach  us  how 
to  proportion  the  best  or  strongest  joint  in  any  given  plates,  as 
no  complete  and  systematic  series  of  tests  has  thus  far  been 
carried  out,  though  such  a  series  has  been  begun  on  the  govern- 
ment testing-machine  at  the  Watertown  Arsenal. 


RIVETED  JOINTS.  $2? 


The  only  tests  to  which  it  seems  to  the  writer  worth  while 
to  make  reference  here  are  : 

i°.  A  portion  of  those  made  by  a  committee  of  the  British 
Institution  of  Mechanical  Engineers,  inasmuch  as,  although  a 
very  large  part  were  made  upon  narrow  strips  with  but  few 
rivets,  nevertheless  a  portion  were  made  upon  wide  strips. 

2°.  The  tests  on  riveted  joints  that  have  been  made  on  the 
government  testing-machine  at  Watertown  Arsenal. 

3°.  A  few  tests  made  by  David  Kirkaldy.  The  following 
are  references  to  these  sets  of  tests,  viz. : 

i°.  The  account  of  this  series  is  to  be  found  at  intervals 
from  1880  to  1885  inclusive,  with  one  supplementary  set  in  1888, 
in  the  proceedings  of  the  British  Institution  of  Mechanical 
Engineers;  but  as  all  except  the  supplementary  set  has  also 
been  published  in  London  Engineering,  these  latter  references 
will  be  given  here  as  follows : 

Engineering  for  1880,  vol.  29,  pages  no,  128,  148,  254,  300,  350. 

"    1881,  vol.  31,      "      427,436,458,508,588. 

"    1885,  vol.  39,      "      524. 

"    1885,  vol.  40,      "      19,43- 
Also,  Proc.  Brit.  Inst.  Mechl.  Engrs.,  Oct.  1888. 

2°.  The  second  series,  referred  to  above,  or  those  made  on 
the  government  testing-machine  at  Watertown  Arsenal,  are  to 
be  found  in  their  reports  as  follows,  viz. : 

Exec.  Doc.     i,  47th  Congress,  2d  Session,  Senate. 

"        5,48th         "          ist       " 
"         "      36,  49th         "  ist       " 

"      31,  49th  2d       "         Housed 

Tests  of  metals  at  Watertown  Arsenal,  Mass.,  June,  1887. 

3°.  The  account  of  Kirkaldy's  tests  is  to  be  found  in  Lon- 
don Engineering,  vol.  xxix.  page  194. 

While  it  is  from  tests  upon  long  joints  that  we  can  derive 


528  APPLIED  MECHANICS. 

correct  and  reliable  information  to  use  in  practice,  and  hence 
while  the  experiments  already  made  give  us  a  considerable 
amount  of  information,  nevertheless  as  the  tests  have  not  yet 
been  carried  far  enough  to  furnish  all  the  information  we  need, 
and  to  settle  cases  that  we  are  liable  to  be  called  upon  to 
decide,  therefore,  before  quoting  the  above  experiments,  some 
of  the  rules  and  proportions  in  use  in  practice  at  the  present 
time,  and  the  modes  of  determining  them,  will  be  first  ex- 
plained. 

In  this  regard  we  must  observe  that  practical  considerations 
render  it  necessary  to  make  the  proportions  different  when  the 
joint  is  in  the  shell  of  a  steam-boiler,  from  the  case  when  it  is 
in  a  girder  or  other  part  of  a  structure. 

In  the  case  of  boiler-work,  the  joint  must  be  steam-tight,  and 
hence  the  pitch  of  the  rivets  must  be  small  enough  to  render 
it  so  :  whereas  in  girder-work  this  requirement  does  not  exist ; 
and  hence  the  pitch  can,  as  far  as  this  requirement  goes,  be 
made  greater. 

It  is  probable,  that,  with  good  workmanship,  we  might  be  able 
to  secure  a  steam-tight  joint  with  considerably  greater  pitches 
than  those  commonly  used  in  boiler-work ;  and  now  and  then 
some  boiler-maker  is  bold  enough  to  attempt  it. 

Tables  of  usual  dimensions  employed  and  recommended  by 
Robert  Wilson,  Thomas  Box,  and  Unwin  respectively,  will  now 
be  given. 

The  following  tables  give  the  proportions  recommended  by 
Robert  Wilson  for  boiler-work,  and  by  Thomas  Box  for  girder- 
work  :  — 


RIVETED  JOINTS. 


529 


PROPORTIONS   GIVEN   BY  ROBERT  WILSON  FOR  USE  IN  BOILER-WORK. 


Diameter  of 

Pitch  of  Rivets. 

Double-Riveted  Butt-joints 

Thickness 
of  Plates. 

Rivets.     All 
Lap-joints  and 
Butt-joints  with 
Single  Strips. 

Pitch  of  Rivets. 
Single-Riveted 
Lap-joint. 

Double-  Rive  ted 
Lap-joints,  and 
Butt-joints  with 
Single  Strips. 

with  Double  Strips. 

Diameter  of 
Rivet. 

Pitch  of  Rivets. 

i 

i 

*ii 

Ta 

1  4 

_ 

_ 

A 

I 

If 

2i 

- 

- 

1 

H 

If 

2i 

f 

2* 

A 

1 

If 

a* 

1 

2f 

i 

3 

4 

It 

*t 

f 

2| 

A 

i 

2t 

2f 

H 

2^ 

1 

i 

2i 

H 

i 

3 

H 

I           . 

2t 

2| 

$ 

3i 

I 

I 

4 

3i 

1 

3i 

H 

I 

2i 

3i 

1 

3i 

1 

I 

H 

3i 

i 

.3f 

II 

If 

2i 

3i 

i 

3t 

1 

It 

*i 

3i 

it 

4 

He  gives,  for  the  lap  in  single  riveting,  3  times  the  diameter 
of  the  rivet,  and  never  more  than  3.3  times. 


PROPORTIONS   RECOMMENDED  BY  THOMAS   BOX  FOR  USE  IN  GIRDER- 
WORK. 
SINGLE-RIVETED  LAP-JOINTS. 


Thickness 
of  Plates. 

Diameter 
of  Rivets. 

Pitch  of 
Rivets. 

Lap. 

i 

A 

It 

j 

A 

iV 

4 

ti 

i 

i 

4 

If 

A 

A 

if 

It 

1 

H 

2 

If 

A 

f 

H 

If 

i 

H 

2^ 

2 

DOUBLE-RIVETED  JOINTS. 


Thickness 
of  Plates. 

Diameter 
of  Rivets. 

Pitch  of 
Rivets. 

Lap. 

| 

-H 

2^ 

2l 

ft 

i 

3 

3 

i 

if 

3rV 

3i 

1*6 

I 

3i 

3l 

« 

rt 

3t 

3^ 

* 

it 

s 

* 

530  APPLIED   MECHANICS. 

i°.  In  regard  to  the  diameter  of  the  rivets,  Robert  Wilson 
first  shows,  that  by  taking  the  crushing-strength  in  front  of  the 
rivet  at  89600,  and  the  shearing-strength  of  the  rivet  at  47040 
Ibs.  per  square  inch,  we  should  find  that  the  joint  would  be  safe 
against  crushing  the  metal  in  front  of  the  rivet  with  a  diameter 
of  rivet  equal  to  2|  times  the  thickness  of  the  plate ;  he  recom- 
mends the  diameters  given  in  the  table,  which  for  plates  \  inch 
and  -j5g  inch  thick  are  double  the  thickness  of  the  plate  (this 
being  a  rule  frequently  used),  and  for  thicker  plates  they  grow 
gradually  less. 

Thomas  Box,  in  deducing  his  diameters  for  girder-work,  gives 
the  formula 

d  =  1 1  +  A, 

from  which  he  calculates  the  diameters  given  in  his  table,  giv- 
ing this  as  an  empirical  rule. 

2°.  In  regard  to  the  pitch  of  the  rivets,  Robert  Wilson,  by 
calling  the  shearing-strength  of  the  rivets  per  square  inch  equal 
to  the  tensile  strength  of  the  plate  per  square  inch,  deduces  the 
formula 

.         ird2    ,      , 

*  =  77  +  * 
/ 

The  values  recommended  by  him  differ,  however,  somewhat 
from  the  results  of  this  formula,  in  order  to  retain  a  larger 
section  of  plate  between  rivets. 

Thomas  Box  deduces  his  values  of  the  pitch  by  considering 
the  shearing-strength  of  the  rivet  per  square  inch  as  equal  to 
|  the  tensile  strength  of  the  plate  per  square  inch,  and  then 
calculating  the  joint  so  as  to  give  equal  strength  for  tearing 
and  shearing. 

3°.  In  regard  to  the  lap,  Robert  Wilson  computes  it  so  that 
there  shall  be  strength  enough  to  resist  breaking  through :  his 
formula  has  been  given  in  §  177.  This  would  give,  for  the  lap, 
the  formula  for  single  riveting, 


RIVE  TED  JOINTS.  5  3 


Hence  he  concludes  that  the  common  rule  of  making  the  dis- 
tance between  the  hole  and  edge  of  the  plate  equal  to  the 
diameter  of  the  rivet  is  to  provide  sufficient  resistance  in  this 
regard  :  this  rule,  however,  he  contradicts  by  his  empirical  rule 
given  just  after  the  table. 

Thomas  Box,  on  the  other  hand,  gives  a  graphical  construc- 
tion for  the  lap,  which  practically  accounts  the  shearing  and  the 
tensile  strength  of  the  plate  per  square  inch  the  same,  and  pro- 
vides sufficient  strength  to  prevent  the  rivet  from  shearing  out 
the  plate  in  front  of  it.  His  results  are  a  little  larger  than 
three  times  the  diameter  of  the  rivet. 

Next,  as  to  Unwin's  recommendations,  — 

i°.  In  regard  to  the  diameter  of  rivet,  he  says  that  the 
diameters  used  in  practice  range  from 


to 


and  recommends,  as  a  good  rule  to  follow, 


It  will  be  seen  that  he  is  thus  recommending  a  little  larger 
diameters  than  Wilson  or  Box. 

2°.  As  to  the  pitch,  he  determines  it  from  the  formula 

(p-  d)tft  =  ^d*fs, 
4 

where/  =  pitch,  /  =  thickness  of  plate,  d  =  diameter  of  rivet, 
ft  =  tensile  strength  of  plate  between  rivet-holes  after  the  rivet- 
ing has  been  done,/^  =  shearing-strength  of  rivet  per  square 
inch,  using  such  values  of  ft  and  fs  as  he  considers  suitable. 
The  result  of  all  this  will  be  shown  in  the  following  tables, 
given  by  Unwin. 


532 


APPLIED   MECHANICS. 


Ratio  of  Tearing  and  Shearing  Strength  j  in  Riveted  Joints,  ft  being  Tensile 
Strength  per  Square  Inch  after  the  Plate  has  been  Punched  or  Drilled. 


Iron  Plates, 

Iron  Rivets. 

Steel  Plates, 

Steel  Rivets. 

Drilled  or 

Drilled  or 

Punched, 

Punched, 

and  An- 

Punched. 

and  An- 

Punched. 

nealed  or 

nealed  or 

Rymered. 

Rymered. 

Single-riveted    .'    

O.Q4 

O.77 

1.26 

i.ot; 

Double-riveted  

I.  O2 

0.8  C 

1.74 

1.17 

Treble-riveted    

1.05 

I.36 

Values  of  Pitch  for  Single  Riveting  for  Various  Values  o 


Single  Riveting. 

Iron  Rivets. 

Steel  Rivets. 

w 

«? 

£ 

> 

I 

22 

"8 

2 

Iron  Drilled  or 

Steel  Punched 

S 
I 

1 

1 

Iron  Punched 
Plates. 

Punched,  and 
Annealed  or 
Rymered 

Steel  Punched 
Plates. 

and  Annealed, 
or  Rymered 
Plates. 

S 

Q 

"rt 

e  : 

Plates. 

| 

8 

Pitch  for  values  of  *—  = 

' 

f* 

0.75      0.85      0.95      1.0        1.05      1.15      1.25      1.35 

•fs 

H 

0.72 

2.46 

2.25 

2.09 

2.O2 

I.96 

1.85 

1.77 

r.69 

1 

I 

0.78 

2.48 

2.28 

2.12 

2.06 

1.99 

1.89 

1.81 

1.72 

* 

tt' 

0.85 

2.58 

2.38 

2.22 

2.15 

2.09 

1.98 

1.90 

1.81 

I 

1 

0.92 

2.69 

2.48 

2.32 

2.25 

2.19 

2.08 

2.00 

1.90 

1 

ii 

0.98 

2.69 

2.40 

2.25 

2.19 

2.13 

2.O3 

'•9S 

1.87 

t 

*tV 

I.IO 

2.79 

2.59 

2-43 

2-37 

2.3I 

2.  2O 

2.12 

2.04 

I 

if 

1.17 

2.8l 

2.62 

2.46 

2.40 

2-34 

2.24 

2.16 

2.08 

i 

if 

1.30 

3-07 

2.86 

2.70 

2.63 

2.56 

2-45 

2.36 

2.28 

1 

RIVETED  JOINTS. 


533 


By  "  real  diameter  "  he  means  the  diameter  after  riveting. 


Double  Riveting. 

jj 

Iron  Rivets. 

Steel  Rivets. 

i 

d 

1 

• 

Iron 

Steel 

K 

z 

'o 

Iron 

Drilled  or 

Steel 

Drilled  or 

"o 

«J 

S3 

Punched 

Punched, 

Punched 

Punched, 

1 

a 

a 

.2 

1 

Plates. 

and  Ry- 

Plates. 

and  Ry- 

Q     • 

1 

mered  Plates. 

mered  Plates. 

a 

"(3 

S 

Q 

H 

'§ 

1 

55 

Pitch  of  rivets  for  value  '—  = 

0.85                  1.00                1.  10               1.20                  1.35 

A 

ii 

0.72 

3-78 

3-33 

3.12 

2.9I 

2.66 

f 

f 

0.78 

3-78 

3-33 

3.12 

2.9I 

2.66 

t^ 

it 

0.85 

3-91 

345 

3-24 

3-03 

2.77 

i 

| 

0.92 

4-05 

3.58 

3-37 

3-16 

2.88 

8 

if 

0.98 

3.82 

3-39 

3.18 

3.00 

2.76 

f 

IT6 

I.  TO 

4.08 

3-63 

342 

3-22 

2.98 

| 

rt 

I.I7 

4.06 

3-63 

342 

3-23 

2.99 

I 

ii 

1.30 

4.42 

3-95 

3-74 

3-52 

3-26 

It  will  be  noticed,  that,  having  used  a  larger  rivet  than  Wil- 
son or  Box,  he  naturally  used  a  larger  pitch. 

For  lap,  he  gives  the  following  values,  computed  by  the  same 
rule  as  Wilson  computes  his,  but  with  a  different  constant ;  and 
he  then  compares  them  with  values  of  i.5</,  which  he  states  to 
be  an  ordinary  rule. 


d 

i 

f 

1 

i 

J 

ii 

ii 

For  iron,  /  = 

I.OO 

1.14 

1.29 

I.4I 

i-55 

1.67 

i.  80 

For  steel,  /  = 

0.86 

0.98 

1.  12 

1.22 

i-35 

1.46 

i-57 

i.& 

0.75 

0.94 

1.  12 

I-3I 

1.50 

1.69 

1.88 

534 


APPLIED  MECHANICS. 


Unwin  gives  as  his  formula  for  computing  the  efficiency, 
i.e.,  the  ratio  of  the  strength  of  the  joint  to  that  of  the  solid 
plate, 

rf  =  K '  ~p~* 

\  ''  j 

where  77  —  efficiency,/  —  pitch,  </—  diameter,  and  K\s  a  con- 
stant. He  then  gives  for  the  efficiencies  that  should  be  ex- 
pected the  following  tables : 


•o 

1 

Single  Riveting. 

1 

<u 

K 

Iron  Drilled  or 

Steel  Punched 

s 

<u 
g  ^ 

'o 

Iron  Punched 

Punched,  and 

Steel  Punched 

and  Annealed, 

'o 

.2  « 

5 

Plates. 

Annealed  or 

Plates. 

or  Rymered 

I 

Q2 

1 

Rymered  Plates. 

Plates. 

0 

§ 

Q 

g 

•a 

Efficiency  of  joints  for  values  of  K  = 

* 

H 

0.77                  0.88                   0.9                   i.o 

^ 

H 

0.72 

0-55 

0.52 

0.58 

0.56 

0-57 

o-55 

0-59 

0-57 

i 

f 

0.78 

0-53 

0.51 

o-55 

0.54 

0-55 

o-53 

0-57 

o-55 

•h 

fi 

0.85 

0.52 

0.49 

o-55 

0.54 

°-53 

0.51 

0-55 

°53 

\ 

j 

0.92 

0.51 

0.49 

0.52 

0.52 

0.52 

0.50 

0-54 

0.52 

£ 

H 

0.98 

,048 

o-45 

0.49 

0.48 

0.49 

0.47 

0.50 

0.48 

1 

ITLB 

1.  10 

0.47 

0.44 

0.48 

0.47 

0.47 

0-45 

0.48 

0.46 

^ 

8 

ii 

I.I7 

0-45 

0.42 

0.46 

0.45 

0-45 

0-43 

0.46 

0.44 

I 

ii 

I.30 

0.42 

0.40 

0.46 

0.45 

0-45 

0-43 

0-45 

0-43 

RIVETED  JOINTS. 


535 


Double  Riveting. 

' 

w 

K 

1 

Iron  Drilled 

Steel  Drillel 

a 

5 

Iron 

or  Punched, 

Steel           or  Punched, 

i 

Punched 

and 

Punched               and 

"o 

u 

u 

Plates. 

Ryniered 

Plates. 

Rymered 

1 

B 

rt 

E 

Plates. 

Plates. 

fl 

P 

is 

P 

H 

c 

1 

Efficiency  of  joints  for  values  of  fT— 

<£ 

0.85             0.95            i.  oo            i.  oo             i.  06 

ft 
T6" 

H 

0.72 

0.69 

0.74 

0.77 

0-75 

0.77 

I 

I 

0.78 

0.68 

0-73 

0-75 

0.73 

o-75 

•TV 

11 

0.85 

0.66 

0.71 

0.74 

0.72 

0.74 

i 

i 

0.92 

0.65 

0.70 

0-73 

0.71 

0.73 

1 

l£ 

0.98 

0.63 

0.67 

0.69 

0.67 

0.69 

i 

'A 

I.IO 

0.62 

0.66 

0.68 

0.66 

0.68 

1 

^ 

I.I7 

0.60 

0.65 

0.66 

0.64 

0.65 

i 

:i 

1.30 

0.60 

0.63 

0.65 

0.63 

0.64 

Punching  and  Drilling. — One  matter  in  this  connection 
that  has  occupied  a  good  deal  of  attention  is  the  relative  advan- 
tages of  punched  and  drilled  holes.  Punching  is  even  now  the 
most  common  practice,  as  it  is  less  expensive  than  drilling. 
In  this  regard  it  should  be  said : 

i°.  When  the  holes  are  drilled,  and  hence  no  injury  is  done 
to  the  metal  between  the  rivet-holes,  this  portion  of  the  plate 
comes  to  have  the  properties  of  a  grooved  specimen,  and  hence 
has  a  greater  tensile  strength  per  square  inch  than  a  straight 
specimen  of  the  same  plate,  as  the  metal  around  the  holes  has 
not  a  chance  to  stretch.  This  excess  tenacity  may  amount 
to  as  much  as  25  per  cent  in  some  cases,  though  it  is  usually 
nearer  10  or  12  per  cent,  depending  not  only  on  the  nature  of 
the  material,  but  also  on  the  proportions. 

2°.  When  the  holes  are  punched,  we  have,  again,  a  grooved 


536  APPLIED  MECHANICS. 

specimen,  but  the  punching  injures  the  metal  around  the  hole, 
and  this  injury  is  greater  the  less  the  ductility  of  the  metal : 
thus,  much  less  injury  is  done  by  the  punch  to  soft-steel  plates 
than  to  wrought-iron  ones,  and  less  to  thin  than  to  thick  plates. 
This  injury  may  reach  as  much  as  35  per  cent,  or  it  may  be 
very  small.  Besides  this,  in  punching  there  is  liability  of  crack- 
ing the  plate,  and  of  not  having  the  holes  in  the  two  plates  that 
are  to  be  united  come  exactly  opposite  each  other.  A  number  of 
tests  on  the  tenacity  of  punched  and  drilled  plates  of  wrought- 
iron,  and  of  mild  steel,  made  on  the  government  testing-machine 
at  Watertown  Arsenal,  are  given  on  page  548  et  seq. 

The  hardening  of  the  metal  by  punching  also  decreases  the 
ductility  of  the  piece. 

The  injury  done  by  punching  may  be  almost  entirely  re- 
moved in  either  of  the  following  ways  :  — 

i°.  By  annealing  the  plate. 

2°.  By  reaming  out  the  injured  portion  of  the  metal  around 
the  hole ;  i.e.,  by  punching  the  hole  a  little  smaller  than  is  de- 
sired, and  then  reaming  it  out  to  the  required  size. 

There  is  a  certain  friction  developed  by  the  contraction  of 
the  rivets  in  cooling,  tending  to  resist  the  giving-way  of  the 
joint ;  but  this  is  likely  to  disappear  before  fracture  takes  place, 
and  cannot,  therefore,  be  depended  upon. 

The  shearing-strength  of  the  rivets  would  appear  to  be  at 
most  about  £  or  £  the  tensile  strength  of  the  plate. 

TESTS    OF    THE    COMMITTEE    OF    THE    BRITISH    INSTITUTION    OF 
MECHANICAL    ENGINEERS. 

The  Committee  on  Riveted  Joints  of  the  British  Institution 
of  Mechanical  Engineers  consisted  of  Messrs.  W.  Boyd,  W.  S. 
Hall,  A.  B.  W.  Kennedy,  R.  V.  J.  Knight,  W.  Parker,  R.  H. 
Twedell,  and  W.  C.  Unwin. 


RIVETED  JOINTS.  537 


Before  beginning  operations  Prof.  Unwin  was  asked  to 
prepare  a  preliminary  report,  giving  a  summary  of  what  had 
already  been  done  by  way  of  experiment,  and  also  to  make 
recommendations  as  to  the  course  to  be  pursued  in  the  tests. 

This  preliminary  report  is  contained  in  vol.  xxix.  of  Engineer- 
ing, on  the  pages  already  cited.  In  regard  to  its  recommenda- 
tions it  is  unnecessary  to  speak  here,  as  the  records  of  the  tests 
show  what  was  done ;  but  in  regard  to  the  summary  of  what 
had  been  done,  it  may  be  well  to  say  that  he  gives  a  list  of 
forty  references  to  tests  that  had  been  made  before  1880,  be- 
ginning with  those  of  Fairbairn  in  1850,  and  ending  with  some 
made  by  Greig  and  Eyth  in  1879,  together  with  a  brief  account 
of  a  number  of  them. 

Almost  all  of  this  work  was  done,  however,  with  small  strips 
with  but  few  rivets,  and  will  not  be  mentioned  here.  Inas- 
much, however,  as  Fairbairn's  proportional  numbers  have  been 
very  extensively  published,  and  are  constantly  referred  to  by 
the  books  and  by  engineers,  it  may  be  well  to  quote  a  portion 
of  what  Unwin  says  in  that  regard,  as  follows  : 

"  The  earliest  published  experiments  on  riveted  joints,  and 
probably  the  first  experiments  on  the  strength  of  riveting  ever 
made,  are  contained  in  the  memoir  by  Sir  Wm.  Fairbairn  in  the 
Transactions  of  the  Royal  Society. 

"  The  author  first  determined  the  tenacity  of  the  iron,  and 
found,  for  the  kinds  of  iron  experimented  upon,  a  mean  tenacity 
of  22.5  tons  per  square  inch  with  the  stress  applied  in  the 
direction  of  the  fibre,  and  23  with  the  stress  across  it.  That 
the  plates  were  found  stronger  in  a  direction  at  right  angles  to 
that  in  which  they  were  rolled  is  probably  due  to  some  error 
in  marking  the  plates. 

"  Making  certain  empirical  allowances,  Sir  Wm.  Fairbairn 
adopted  the  following  ratios  as  expressing  the  relative  strength 
of  riveted  joints  : 


A  PI3  LI  ED  MECHANICS. 


Solid  plate    ............   100 

Double-riveted  joint  ......     '.     .     .     70 

Single-riveted  joint     .........     50 

These  well-known  ratios  are  quoted  in  most  treatises  on  rivet- 
ing, and  are  still  sometimes  referred  to  as  having  a  considerable 
authority. 

"  It  is  singular,  however,  that  Sir  Wm.  Fairbairn  does  not 
appear  to  have  been  aware  that  the  proportion  of  metal 
punched  out  in  the  line  of  fracture  ought  to  be  different  in 
properly  designed  double  and  single  riveted  joints.  These 
celebrated  ratios  would  therefore  appear  to  rest  on  a  very 
unsatisfactory  analysis  of  the  experiments  on  which  they  are 
based.  Sir  Wm.  Fairbairn  also  gives  a  well-known  table  of 
standard  dimensions  for  riveted  joints.  It  is  not  very  clear 
how  this  table  has  been  computed,  and  it  gives  proportions 
which  make  the  ratio  of  tearing  to  shearing  area  different  for 
different  thicknesses  of  plate.  There  is  no  good  reason  for 
this." 

As  to  the  tests  which  constitute  the  experimental  work  of 
the  committee,  these  were  made  by  or  under  the  direction  of 
Prof.  A.  B.  W.  Kennedy,  of  London.  Steel  plates  and  steel 
rivets  were  used  throughout,  the  steel  containing  about  0.18  per 
cent  of  carbon,  and  having  a  tensile  strength  varying  from 
about  62000  to  about  70000  pounds  per  square  inch,  and  hence 
being  a  little  harder  than  would  correspond  to  our  American 
ideas  of  what  is  suitable  for  use  in  steam-boilers.  The  greater 
portion  of  the  work  was  performed  by  the  use  of  a  testing- 
machine  of  100000  pounds  capacity,  and  hence  one  which  did 
not  admit  of  testing  wide  strips  with  a  sufficient  number  of 
rivets  to  correspond  to  the  cases  which  occur  in  practice; 
indeed,  only  eighteen  of  the  tests  were  made  on  such  strips. 
Nevertheless,  a  brief  summary  of  what  was  done  will  be  given 
here,  though  some  of  the  conclusions  which  he  drew  are  al- 


RIVETED  JOINTS.  539 


ready,  and  others  are  liable  to  be,  proved  untrue  by  tests  of 
wide  strips.  The  tests  made  by  Prof.  Kennedy  up  to  1885 
consisted  of  fourteen  series  numbered  I  to  V,  VA  and  VI  to 
XIII,  and  covering  290  experiments,  64  on  punched  or  drilled 
plates,  97  on  joints,  44  on  the  tenacity  of  the  plates  used  in 
the  joints,  33  on  the  tenacity  and  shearing-resistance  of  the 
rivet-steel  used  in  the  joints,  and  the  remaining  52  on  various 
other  matters. 

The  first  three  series  were  upon  the  tenacity  of  the  steel 
used,  and  showed  it  to  be,  as  stated,  from  62000  to  70000  pounds 
per  square  inch,  with  an  ultimate  elongation  of  23  to  25  per 
cent  in  a  gauged  length  of  ten  inches ;  the  tenacity  of  the 
rivet-steel  being  practically  the  same  as  that  of  the  plates. 
The  fourth  series  showed  the  shearing-strength  of  the  rivet- 
steel  to  be  about  55000  pounds  per  square  inch  when  tested  in 
one  way,  and  59000  pounds  per  square  inch  when  tested  in 
another  way  which  corresponded,  as  Kennedy  claims,  better 
to  the  conditions  of  a  rivet,  though  neither  was  by  using  a 
riveted  joint. 

The  tests  of  series  V  and  VA  were  made  upon  pieces  of 
plate  which  had  been  punched  or  drilled,  in  other  words,  on 
grooved  specimens ;  and,  as  might  be  expected,  these  specimens 
showed  invariably  an  increase  in  tensile  strength  over  the 
straight  specimens.  In  the  J"  and  f  «"  plates  drilled  with  holes 
i  inch  in  diameter  and  2  inches  pitch,  the  net  metal  between 
the  holes  had  a  tenacity  II  to  12  per  cent  greater  than  that  of 
the  untouched  plate.  Even  with  punched  holes  the  metal  had 
a  similar  excess  of  tenacity  of  over  6  per  cent.  The  remaining 
eight  series,  VI  to  XIII  inclusive,  were  made  on  riveted  joints, 
the  first  five  on  single-riveted  lap-joints,  and  the  last  three, 
or  XI,  XII,  and  XIII,  on  double-riveted  lap  and  butt  joints. 

Series  VI  was  made  on  twelve  joints  in  -f-inch  plates  which 
contained  only  two  rivets  each,  the  proportions  not  being  in- 
tended to  be  those  of  practice,  but  such  as  should  give,  to 


54°  APPLIED   MECHANICS. 


some  extent,  limiting  values  for  the  resistances  of  the  plate  to 
tearing,  and  of  the  rivets  to  shearing  and  pressure.  The  results 
were  rather  irregular;  and  the  main  conclusion  which  he  drew, 
was,  that  if  the  joint  is  not  to  break  by  shearing,  the  ratio  of 
the  tearing  to  the  shearing  area  must  be  computed  on  a  much 
lower  value  of  shearing-strength  per  square  inch  than  the  ex- 
periments of  series  IV  had  shown;  indeed,  some  of  the  joints 
of  series  VI  gave  way  by  shearing  the  rivets  at  loads  no  greater 
than  36000  pounds  per  square  inch  of  shearing-area. 

Series  VII  was  made  upon  six  (single-riveted  lap)  joints  in 
-f-inch  plate,  with  only  three  f-inch  rivets  in  each  joint,  and 
with  varying  pitch  and  lap  ;  all  these  joints  breaking  by  shear- 
ing the  rivets.  His  conclusion  from  these  tests  was,  that  the 
lap  need  not  be  more  than  1.5  times  the  diameter  of  the  rivet. 

Series  VIII  was  made  on  eighteen  (single-riveted  lap)  joints 
in  six  sets  of  three  each,  and  these  are  the  only  single-riveted 
lap-joints  which  he  tested,  having  as  many  as  seven  rivets  each. 
The  results  are  given  in  the  accompanying  table. 

Before  giving  the  table,  it  may  be  said  that  No.  652  was  in- 
tended to  have  such  proportions  as  to  be  equally  likely  to  give 
way  by  tearing  or  by  shearing,  the  intensity  of  the  shearing- 
strength  being  assumed  as  two-thirds  that  of  the  tensile 
strength  of  the  steel,  while  the  bearing-pressure  per  square 
inch  was  intended  to  be  about  7.5  per  cent  greater  than  the 
tension.  No.  653  was  proportioned  with  excess  of  shearing  or 
rivet-area,  No.  654  with  defect  of  shearing-area,  No.  655  with 
excess  of  tearing  or  plate  area,  No.  656  with  defect  of  tearing- 
area,  and  No.  657  with  excess  of  bearing-pressure,  the  different 
proportions  being  arrived  at  by  varying  the  pitch  and  diameter 
of  the  rivets,  and,  in  the  case  of  657,  the  thickness  of  the  plate 
also.  The  margin  (or  lap  minus  radius  of  rivet)  was  £  inch  in 
each  case.  The  following  table  will  show  how  far  these  inten- 
tions were  realized,  and  further  comments  will  be  deferred  till 
later. 


RIVETED  JOINTS. 


541 


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542  APPLIED   MECHANICS. 

Series  IX  was  made  on  twenty-one  joints  in  f-inch  plate 

(each  containing  only  two  rivets)  designed  in  a  manner  similar 

to  series  VIII,  while  three  were  afterwards  made  from  some 

*of  the   broken  plates,  with  as  heavy  rivets  as  it  was  deemed 

possible  to  make  tight. 

From  these  tests  Kennedy  thinks  it  fair  to  conclude  — 

I  °.  That  the  efficiency  of  a  single-riveted  lap-joint  in  a  |-inch 
plate  cannot  be  greater  than  50  per  cent,  unless  rivets  larger 
than  i.i  inch  are  used;  and  he  also  calls  attention  to  the  fact 
that,  as  he  claims,  strength  is  gained  by  putting  more  metal  in 
the  heads  and  ends  of  the  rivets,  claiming  that  it  will  make 
also  a  tighter  joint  for  boiler-work. 

Series  X  was  made  on  eight  single-riveted  lap-joints  in 
^inch  and  f-inch  plate,  made  from  the  broken  specimens  of 
series  V  and  VA ;  they  also  had  only  two  rivets  each.  These 
joints  were  made  with  a  view  of  investigating  the  effect  of 
more  or  less  bearing-pressure.  He  claims  that  high  bearing- 
pressure  induces  a  low  shearing-strength  in  the  rivets,  and  that 
the  bearing-pressure  should  not  exceed  about  96000  pounds  per 
square  inch  ;  also,  that  when  a  large  bearing-pressure  is  used, 
the  "  margin  "  should  be  extra  large  to  prevent  distortion,  and 
consequent  local  inequalities  of  stress  ;  also,  that  smaller  bearing- 
pressures  do  not  much  affect  the  strength  of  the  joint  one  way 
or  the  other. 

Series  XI  was  made  upon  twelve  specimens  of  double-riveted 
joints ;  three  being  lap-joints  in  f-inch  plate,  three  lap-joints  in 
J-inch  plate,  three  butt-joints  with  two  equal  covers  in  f-inch 
plate,  and  three  butt-joints  with  two  equal  covers  in  f-inch 
plate.  Kennedy  designed  these  joints  with  a  view  to  their 
being  equally  likely  to  fail  by  tearing  or  by  shearing.  His  as- 
sumptions and  the  results  of  the  tests  are  all  given  in  the  fol- 
lowing table : 


RIVETED  JOINTS. 


543 


SERIES  XL     DOUBLE-RIVETED  LAP  AND  BUTT  JOINTS— AVERAGES. 


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Series  XII  contains  the  same  joints  as  series  XI,  the  strained 
ends  having  been  cut  off,  and  the  rest  redrilled  and  riveted  by 
means  of  Mr.  Twedell's  hydraulic  riveter;  and  series  XIII  con- 
tained the  same  joints  treated  a  second  time  in  the  same  way. 
These  experiments,  so  far  as  they  went,  showed  no  gain  in 
ultimate  strength  to  result  from  hydraulic  as  compared  with 
hand-riveting ;  but  it  was  found  that,  through  a  misunderstand- 
ing, they  had  been  riveted  up  at  a  pressure  much  lower  than 
that  intended  by  Mr.  Twedell. 

On  the  other  hand,  the  load  at  which  visible  slips  occurred 
was  about  twice  as  much  greater  with  hydraulic  as  with  hand 
riveting. 


544  APPLIED   MECHANICS. 


KENNEDY  S    CONCLUSIONS. 

The  following  are  a  portion  of  what  he  gives  as  his  con- 
clusions : 

i°.  The  metal  between  the  rivet-holes  had  a  considerably 
greater  tensile  resistance  per  square  inch  than  the  unperfo- 
rated  metal. 

2°.  In  single-riveted  joints,  with  the  metal  that  he  used,  he 
assumed  about  22  tons  (49280  Ibs.)  per  square  inch  as  the  shear- 
ing-strength of  the  rivet-steel  when  the  bearing-pressure  is 
below  40  tons  (8960x3  Ibs.)  per  square  inch.  In  double-riveted 
joints  with  rivets  of  about  f-inch  diameter  we  can  generally 
assume  24  tons  (53760  Ibs.)  per  square  inch,  though  some  fell 
to  22  tons  (49280  Ibs.). 

3°.   He  advises  large  rivet  heads  and  ends. 

4°.  For  ordinary  joints  the  bearing-pressure  should  not  ex- 
ceed 42  or  43  tons  (94000  or  96000  Ibs.)  per  square  inch.  For 
double-riveted  butt-joints  a  higher  bearing-pressure  may  be 
allowed ;  the  effect  of  a  high  bearing-pressure  is  to  lower  the 
shearing-strength  of  the  steel  rivets. 

5°.  He  advises  for  margin  the  diameter  of  the  hole,  except 
in  double-riveted  butt-joints,  where  it  should  be  somewhat 
larger. 

6°.  In  a  double-riveted  butt-joint  the  net  metal,  measured 
zigzag,  should  be  from  30  to  35  per  cent  greater  than  that  meas- 

2          d 
ured  straight  across,  i.e.,  the  diagonal  pitch  should  be  — /  -j — , 

where/  =  transverse  pitch  and  d  =  diameter  of  rivet-hole. 

7°.  Visible  slip  occurs  at  a  point  far  below  the  breaking- 
load,  and  in  no  way  proportional  to  that  load. 

Kennedy  thinks  that  these  tests  enable  him  to  deduce  rules 
for  proportioning  riveted  joints,  and  the  following  are  his  rules, 


RIVETED  JOINTS.  545 


(a)  For  single-riveted  lap-joints  the  diameter  of  the  hole 
should  be  2\  times  the  thickness  of  the  plate,  and  the  pitch  of 
the  rivets  2f  times  the  diameter  of  the  hole,  the  plate-area  being 
thus  71  per  cent  of  the  rivet-area.  If  smaller  rivets  are  used, 
as  is  generally  the  case,  he  recommends  the  use  of  the  follow- 
ing formula : 

d* 
P  —  &  ~j  +  ^> 

where  /  =  thickness  of  plate,  d  =  diameter  of  rivet,  and  /  = 
pitch. 

For  30-ton  (67200  Ibs.)  plate,  and  22-ton  (49280  Ibs.)  rivets,  a  —  0.524 
For  28-ton  (62720  Ibs.)  plate,  and  22-ton  (49280  Ibs.)  rivets,  a  —  0.558 
For  30-ton  (67200  Ibs.)  plate,  and  24-ton  (53760  Ibs.)  rivets,  a  =  0.570 
For  28-ton  (62720  Ibs.)  plate,  and  24  ton  (53760  Ibs.)  rivets,  a  =  0.606 

Or,  as  a  mean,  a  =  0.56. 

(&)  For  double-riveted  lap-joints  he  claims  that  it  would  be 
desirable  to  have  the  diameter  of  the  rivet  2\  times  the  thick- 
ness of  the  plate,  and  that  the  ratio  of  pitch  to  diameter  of 
hole  should  be  3.64  for  3O-ton  (67200  Ibs.)  plate  and  22-ton 
(49280  Ibs.)  or  24-ton  (53760  Ibs.)  rivets,  and  3.82  for  28-ton 
(62720  Ibs.)  plate. 

Here,  however,  it  is  specially  likely  that  this  size  of  rivet 
may  be  inconveniently  large,  and  then  he  says  they  should  be 
made  as  large  as  possible,  and  the  pitch  should  be  determined 
from  the  formula  to 

/  =  «y  +  < 

where, 

For  30-ton  (67200  Ibs.)  plate,  and  24-ton  (53760  Ibs.)  rivets,  a  =  1.16 
For  28-ton  (62720  Ibs.)  plate,  and  22-ton  (49280  Ibs.)  rivets,  a  =  1.16 
For  3o-ton  (67200  Ibs.)  plate,  and  22-ton  (49280  Ibs.)  rivets,  a  =  1.06 
For  28-ton  (62720  Ibs.)  plate,  and  24-ton  (53760  Ibs.)  rivets,  a  =  1.24 


54-6  APPLIED   MECHANICS. 

(c)  For  double-riveted  butt-joints  he  recommends  that  the 
diameter  of  the  hole  should  be  about  1.8  times  the  thickness 
of  the  plate,  and  the  pitch  4.1  times  the  diameter  of  the  hole, 
and  that  this  latter  ratio  be  maintained  even  when  the  former 
cannot  be. 

Two  of  the  principal  participants  in  the  discussion  of  the 
report  were  Mr.  R.  Charles  Longridge  and  Prof.  W.  C.  Unwin. 

Mr.  Longridge  was  of  the  opinion  that  wider  strips  with 
more  rivets  should  have  been  used  ;  that  holding  the  specimens 
in  the  machine  by  means-  of  a  central  pin  at  each  end  was  not 
the  best  method ;  that  the  results  obtained  from  specimens 
which  had  been  made  from  the  remnants  of  other  fractured 
specimens  were  at  least  questionable,  for,  even  if  the  plate  had 
not  been  injured,  the  ratio  of  the  length  to  the  width  of  the 
narrowest  part  was  different  after  the  strained  ends  were  cut 
off  from  what  it  was  before  ;  that  machine-riveting  should  have 
been  adopted  throughout  instead  of  hand-riveting,  as  it  is  not 
possible  to  secure  uniformity  with  the  latter  even  were  it  all 
done  by  the  same  man,  as  he  would  be  more  tired  at  one  time 
than  at  another ;  that  experiments  should  be  made  to  determine 
the  effect  of  different  sizes  and  different  shapes  of  heads,  as 
well  as  of  different  pressures  upon  the  load  causing  visible  slip  ; 
and  that  experiments  should  be  made  upon  chain-riveting,  as 
he  thought  the  chain-riveted  joint  would  show  a  greater  effi- 
ciency than  the  staggered. 

Professor  Unwin  said  : 

i°.  In  examining  the  results  to  ascertain  how  far  a  variation 
from  the  best  proportions  was  likely  to  affect  the  strength  of 
the  joint,  he  found  that  while  the  ratio  of  rivet  diameter  to 
thickness  of  plate  varied  21  per  cent,  the  ratio  of  shearing  to 
tearing  area  30  per  cent,  and  the  ratio  of  crushing  to  tearing 
area  34  per  cent,  the  efficiency  of  the  weakest  joint  was  only 
six  per  cent  less  than  that  of  the  strongest,  or,  in  other  words, 


RIVETED  JOINTS.  S47 


the  whole  variation  of  strength  was  only  1 1  per  cent  of  the 
strength  of  the  weakest  joint. 

2°.  With  reference  to  the  effect  which  the  crushing-pres- 
sure on  the  rivet  produced  upon  the  strength  of  the  joint, 
there  were  some  old  experiments,  which  showed  that,  when 
the  bearing-pressure  on  the  rivet  became  very  large  there  was 
a  great  diminution  in  the  apparent  tenacity  of  the  plate  in 
the  case  of  riveted  joints  in  iron.  Why  should  the  crushing- 
pressure  affect  either  the  tenacity  of  the  plate  or  the  shearing 
resistance  of  the  rivet?  He  believed  that  it  did  not  really 
affect  either.  What  happened  was  that,  if  the  crushing-pres- 
sure exceeded  a  certain  limit,  there  was  a  flow  of  the  metal, 
and  the  section  which  was  resisting  the  load  was  diminished. 
Either  the  section  of  the  plate  in  front  of  the  rivet,  if  the  plate 
was  soft,  or  the  section  of  the  rivet  itself,  if  the  rivet  was  soft, 
became  reduced. 

3°.  He  thought  that  the  point  at  which  visible  slip  began 
was  the  initial  point  at  which  the  friction  of  the  plates  was 
overcome,  and  of  course  was  greater  the  greater  the  grip 
upon  the  plates,  and  hence  greater  in  machine  than  in  hand 
riveting.  In  some  cases  with  hydraulic  riveting  loads  were  got 
as  high  as  10  tons  (22400  Ibs.)  per  square  inch  of  rivet  section 
before  slipping  began. 

4°.  In  regard  to  the  rules  for  proportioning  riveted  joints, 
he  preferred  to  distinguish  the  joints  as  single-shear  and  double- 
shear  joints,  and  then  we  have  the  following  three  equations : 
one  by  equating  the  load  to  the  tearing-resistance  of  the  plates, 
a  second  by  equating  it  to  the  shearing-resistance  of  the  rivets, 
and  a  third  by  equating  it  to  the  crushing  resistance ;  these 
three  determining  the  thickness  of  the]  plate,  the  diameter  of 
the  rivet,  and  the  pitch. 

By  taking  the  crushing  as  double  the  tenacity,  we  should 
obtain  for  single  shear  d  =  2.57^,  and  for  double-shear,  d  = 

I.27/. 


548  APPLIED  MECHANICS. 

In  a  single-shear  joint  the  rivet  cannot  generally  be  made 
so  big,  and  in  the  double-shear  it  could  not  always  be  made  so 
small,  hence  the  rivet  diameter  is  chosen  arbitrarily,  and  then 
the  single-shear  joint  is  proportioned  by  the  equations  for  shear- 
ing and  tearing,  no  attention  being  paid  to  the  crushing,  while 
the  double-shear  joint  is  proportioned  by  the  equations  for 
crushing  and  tearing,  no  attention  being  paid  to  the  shearing. 

5°.  The  general  drift  of  the  report  was  to  advocate  the  use 
of  larger  rivets.  Whether  this  could  be  done  or  not,  he  could 
not  say.  For  lap-joints  it  would  increase  the  strength,  whereas 
for  double-shear  joints  he  was  not  sure  that  it  would  not  be 
better  to  diminish  the  size  of  the  rivet,  and  hence  the  crushing- 
pressure. 

This  report  has  been  given  so  fully  because  it  emanates 
from  a  committee  of  the  British  Institution  of  Mechanical 
Engineers;  but  inasmuch  as  series  VIII  is  the  only  one  where 
wide  strips  were  used,  it  seems  to  the  writer  that  any  conclu- 
sions which  may  be  drawn  from  any  of  the  other  tests  given 
in  the  report  require  confirmation  by  tests  on  wide  strips  with 
more  rivets,  before  being  accepted  as  true. 

Government  Experiments. — The  references  to  these  experi- 
ments are  the  following : 

Executive  Document  No.  i,  47th  Congress,  26.  session,  Senate. 
Executive  Document  No.  5,  48th  Congress,  ist  session,  Senate. 
Executive  Document  No.  36,  49th  Congress,  ist  session,  Senate. 
Executive  Document  No.  31,  49th  Congress,  2d  session,  House, 

Part  II. 
Executive  Document  No.  16,  5oth  Congress,  ist  session,  House. 

Those  that  have  been  published  up  to  the  present  time 
may  be  divided  into  three  parts :  — 

i°.  Those  contained  in  the  first  two  Executive  Documents 
mentioned  above. 

2°.  Those  contained  in  the  third  and  fourth. 


RIVETED  JOINTS.  549 


3°.  Those  contained  in  the  fifth. 

Summaries  of  these  sets  of  tests  will  be  given  here  in  their 
order,  as  each  set  was  made  with  certain  special  objects  in 
view,  and,  if  not  all,  at  any  rate  the  I  °  and  2°,  form,  as  has  been  al- 
ready stated,  the  first  portion  of  a  systematic  series ;  and  it  seems 
to  the  author  that,  although  the  series  are  not  yet  completed, 
yet  these  tests  themselves  furnish  more  reliable  information  in 
regard  to  the  behavior  and  the  strength  of  joints  than  any  other 
experiments  that  have  been  made,  and  that  the  figures  them- 
selves furnish  the  engineer  with  the  means  of  using  his  judg- 
ment in  many  cases  where  he  had  no  reliable  data  before. 

A  perusal  of  the  tables  will  give  a  good  idea  of  the  shear- 
ing-strength per  square  inch  of  the  rivet  iron,  which  is  seen  to 
be  less  than  the  tensile  strength  of  the  solid  plate ;  also  the 
effect  on  strength  of  the  plates  due  to  the  entire  process  of 
riveting,  punching,  drilling,  and  driving  the  rivets ;  also  the 
efficiencies  of  the  joints  tested. 

One  of  the  strongest  single-riveted  joints  tested  was  a  single- 
riveted  lap-joint  with  a  single  covering-strip. 

The  apparent  anomaly  of  the  punched  plates  in  a  few  cases, 
showing  a  greater  strength  than  the  drilled  plates,  is  explained 
by  Mr.  Howard  to  be  due  to  the  strengthening  effect  of  cold- 
punching  combined  with  smallness  of  pitch,  inasmuch  as  then 
the  masses  of  hardened  metal  on  the  two  sides  re-enforce  each 
other. 

Further  than  this,  the  student  is  left  to  study  the  figures 
themselves  as  to  the  effect  of  different  proportions,  etc. 

In  regard  to  the  first  series,  i.e.,  those  contained  in  the  first 
two  Executive  Documents  mentioned,  it  is  stated  in  the  report 
that  — 

i°.  "The  wrought-iron  plate  was  furnished  by  one  maker 
out  of  one  quality  of  stock." 

2°  "  The  steel  plates  were  supplied  from  one  heat,  cast  in 
ingots  of  the  same  size;  the  thin  plates  differing  from  the 


55°  APPLIED   MECHANICS. 

thicker  plates  only  in  the  amount  of  reduction  given  by  the 
rolls." 

The  modulus  of  elasticity  of  the  metal  was,  iron  platey 
31970000  Ibs. ;  steel  plate,  28570000  Ibs. 

In  the  tabulated  results,  the  manner  of  fracture  is  shown 
by  sketches  of  the  joints,  and  is  further  indicated  by  heavy 
figures  in  columns  headed  "  Maximum  Strains  on  Joints,  in  Ibs.,. 
per  Square  Inch." 


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571 


GOVERNMENT   TESTS   OF  GROOVED   SPECIMENS. 


Tensile  Tests  of  4-in. 
Grooved  Specimens 
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40850 

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Tensile  Tests  of  ^-in. 
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Tensile  Tests  of  £-in. 
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59060 

2.49 

0.249 

58100 

2.47 

0.249 

63900 

2-43 

0.250 

61640 

2-95 

0.251 

56530 

3-oi 

0.249 

58780 

3-°4 

0.253 

55500 

2.97 

0.252 

60060 

2.98 

0.251 

54050 

2-97 

0.249 

56040 

Tensile  Tests  of  l-in 
Grooved  Specimens 
Steel  Plate 

Drilled. 

i 

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0.50 

0.247 

65610 

0.51 

0.249 

66370 

0.51 

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67420 

0.52 

0.248 

67750 

0.52 

0.252 

61910 

1.03 

0.247 

67090 

i.  02 

0.250 

66390 

1.02 

0.246 

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0.250 

67730 

1.  01 

0.247 

66020 

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66090 

1.54 

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1.52 

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63350 

1.50 

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64370 

1.54 

0.248 

64895 

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0.252 

64320 

2.00 

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62970 

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59550 

3.02 

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59700 

3-00 

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63370 

3.00 

0.251 

58630 

3-03 

0.252 

63940 

572 


APPLIED   MECHANICS. 


IRON  PUNCHED. 


IRON  DRILLED. 


STEEL  PUNCHED. 


STEEL  DRILLED. 


Tensile  Tests  of 

Grooved  Wrought- 
Iron  Plates. 

1 

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jj 

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£ 

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47000 

0.98 

0.370 

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0.382 

39760 

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0.383 

36630 

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37600 

2.98 

0-395 

36340 

2.98 

0.392 

39210 

3-47 

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37680 

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38340 

0-97 

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1.97 

0.512 

43430 

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2.41 

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38950 

2.90 

0.517 

37290 

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37800 

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37770 

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35730 

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3-50 

0.616 

36940 

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0.619 

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0.98 

0.678 

50840 

I.OI 

0.682 

46590 

1.49 

0.688 

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3.48 

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0.692 

3938o 

Tensile  Tests  of 

Grooved  Wrought- 
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j= 

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0.376 

50870 

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0-377 

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0.380 

49830 

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51240 

1.49 

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1.98 

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0.516 

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0.97 

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0.626 

48350 

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47170 

1.49 

0.629 

46530 

2.98 

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48220 

3-46 

0.616 

47770 

3-47 

0.617 

44900 

3-91 

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44840 

3.96 

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52780 

».$» 

0.692 

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0.700 

47750 

3-49 

0.692 

46350 

Tensile  Tests 

of 

Grooved  Steel  Plates. 

fl 

35 

*£* 

£ 

c 

«     £ 

72 

£ 

llfl 

P 

P 

Inch. 

Inch. 

1.99 

0-365 

61890 

0.99 

0-494 

70080 

I.OO 

0.492 

68130 

1.50 

0-497 

66340 

i-5i 

0-494 

63810 

1.99 

0.499 

55930 

1.97 

0.500 

64260 

2-43 

0.502 

52050 

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0.504 

64360 

3-oo 

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60320 

2-99 

0.503 

62430 

3-50 

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3-50 

0-505 

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0.497 

48010 

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0.618 

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56730 

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54220 

Tensile  Tests 

of 

Grooved  Steel  Plates. 

11 

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Inch. 

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1.97 

0.369 

63620 

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0.498 

66220 

0.99 

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2.03 

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66730 

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0-497 

67950 

2.52 

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67440 

3.01 

0.502 

66310 

3.01 

0.503 

66190 

3-49 

0.504 

64920 

3-5° 

0.502 

65210 

3-99 

0.499 

64470 

4.00 

0.498 

64810 

4.00 

0.503 

64690 

4.00 

0.498 

64140 

0.99 

0.619 

60290 

1.49 

0.614 

63610 

1.49 

0.616 

63450 

2-49 

0.620 

59170 

2.50 

0.619 

59600 

3.01 

0.617 

59270 

3-50 

0.614 

61610 

3-49 

0.617 

62060 

4.00 

0.615 

60330 

4.01 

0.617 

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TENSILE    TESTS  OF  RIVETED  JOINTS.  573 

Next  will  be  given  the  two  series  of  tests  already  referred 
to,  with  Mr.  Howard's  analysis  of  them. 

TENSILE    TESTS    OF    RIVETED    JOINTS. 

"  Earlier  experiments  on  this  subject  made  with  single  and 
double  riveted  lap  and  butt  joints  in  different  thicknesses  of 
iron  and  steel  plate,  together  with  the  tests  of  specimens  pre- 
pared to  illustrate  the  strength  of  constituent  parts  of  joints, 
are  recorded  in  the  report  of  tests  for  1882  and  1883. 

From  the  results  thus  obtained  it  appeared  desirable  to 
institute  a  synthetical  series  of  tests,  beginning  with  the  most 
elementary  forms  of  joints  in  which  the  stresses  are  found  in 
their  least  complicated  state.  To  meet  these  conditions,  a 
series  of  joints  have  been  prepared  which  may  be  designated  as 
single-riveted  butt-joints,  in  which  the  covers  are  extended  so 
as  to  be  grasped  in  the  testing-machine ;  thereby  enabling  one 
plate  of  the  joint  to  be  dispensed  with,  and  securing  the  test  of 
one  line  of  riveting. 

Such  a  joint,  made  with  carefully  annealed  mild  steel  plate 
of  superior  quality,  with  drilled  holes,  seems  well  adapted  to 
demonstrate  the  influence  on  the  tensile  strength  of  the  metal 
taken  across  the  line  of  riveting,  of  variations  in  the  width  of 
the  net  section  between  rivets,  and  variations  in  the  compres- 
sion stress  on  the  bearing-surface  of  the  rivets  ;  elements  which 
are  believed  to  be  fundamental  in  all  riveted  construction. 

This  series  comprises  2 16  specimen  joints,  the  thickness  of  the 
plate  ranging  from  J"  to  |",  advancing  by  eighths.  The  covers 
are  from  -f%"  to  TV'-  The  rivets  are  wrought-iron,  and  range  from 
•£•$"  to  lyV  diameter ;  they  are  machine-driven  in  drilled  holes 
yig-"  larger  in  diameter  than  the  nominal  size  of  the  rivets.  Ten- 
sile tests  of  the  material  accompany  the  tests  of  the  joints. 

From  each  sheet  of  steel  two  test-strips  were  sheared,  one 
lengthwise  and  one  crosswise.  The  strips  were  2\"  wide  and  24" 
to  36"  long;  they  were  annealed  with  the  specimen  plates,and  had 
their  edges  planed,  reducing  their  widths  to  \\"  before  testing. 


5/4  APPLIED  MECHANICS. 

Micrometer  readings  were  taken  in  10"  along  the  middle  of 
the  length  of  each. 

The  strength  and  ductility  appear  to  be  substantially  the 
same  in  each  direction.  But  the  practice  of  the  rolling-mill 
where  these  sheets  were  rolled  is  such  that  nearly  the  same 
amount  of  work  may  have  been  given  the  steel  in  each  direc- 
tion ;  that  is,  lengthwise  and  crosswise  the  finished  sheet. 

The  ingots  of  open-hearth  metal  are  first  rolled  down  to 
slabs  about  6"  thick,  then  reheated  and  rolled  either  length- 
wise or  crosswise  their  former  direction,  as  best  suits  the  re- 
quired finished  dimensions. 

The  tensile  tests  show  among  the  thinner  plates  a  relatively 
high  elastic  limit  as  compared  with  the  tensile  strength  ;  in  the 
-f$"  plate  the  percentage  is  72.2,  while  with  the  {-"  plate  the 
percentage  is  found  to  be  53.3. 

It  is  noticeable  that  the  thinner  plates  particularly  exhibit 
a  large  stretch  immediately  following  the  elastic  limit,  and  the 
stretching  is  continued  at  times  under  a  load  lower  than  that 
which  has  been  previously  sustained.  It  is  characteristic  of  all 
the  thicknesses  that  a  considerable  stretch  takes  place  under 
loads  approaching  the  tensile  strength — in  some  cases  the 
stretch  increases  5  to  6  per  cent,  while  the  stress  advances  1000 
pounds  per  square  inch  or  less.  Herein  is  found  a  valuable 
property  of  this  metal  as  a  material  for  riveted  construction. 
The  stress  from  the  bearing-surface  of  the  rivets  is  distributed 
over  the  net  section  of  plate  between  the  rivets,  due  to  the  large 
stretch  of  the  metal,  with  little  elevation  of  the  stress,  and  a 
nearer  approximation  of  uniform  stress  in  this  section  attained 
than  is  found  in  a  brittle  or  less  ductile  metal.  The  joints  were 
held  for  testing  in  the  hydraulic  jaws  of  the  testing-machine, 
having  24"  exposure  between  them.  A  loose  piece  of  steel 
the  same  thickness  as  the  plate  was  placed  between  the  covers 
to  receive  the  grip  of  the  jaws,  and  avoid  bending  the  covers. 

Elongations  were  measured  in  a  gauged  length  of  5",  the 
micrometer  covering  the  joint  at  the  middle  of  its  width.  Loads 


TENSILE    TESTS  OF  RIVETED  JOINTS.  575 


were  applied  in  increments  of  1000  pounds  per  square  inch  of  the 
gross  section  of  the  plate,  the  effect  of  each  increment  determined 
by  the  micrometer,  and  permanent  sets  observed  at  intervals. 

The  progress  of  the  test  of  a  joint  is  generally  marked 
by  three  well-defined  periods.  In  the  first  period  greatest 
rigidity  is  found,  and  it  is  thought  that  the  joint  is  now  held 
entirely  by  the  friction  of  the  rivet-heads,  and  the  movement  of 
the  joint  is  principally  that  due  to  the  elasticity  of  the  metal. 

The  second  period  is  distinguished  by  a  rapid  increase  in 
the  stretch  of  the  joint ;  attributed  to  the  overcoming  of  the 
friction  under  the  rivet-heads  and  closing  up  any  clearance 
about  the  rivets,  bringing  them  into  bearing  condition  against 
the  fronts  of  the  rivet-holes.  Rivets  which  are  said  to  fill  the 
holes  can  hardly  do  so  completely,  on  account  of  the  contrac- 
tion of  the  metal  of  the  rivet  from  a  higher  temperature  than 
that  of  the  plate,  after  the  rivet  is  driven. 

After  a  brief  interval  the  movement  of  the  joint  is  retarded, 
and  the  third  period  is  reached.  The  stretch  of  the  joint  is 
now  believed  to  be  due  to  the  distortion  of  the  rivet-holes  and 
the  rivets  themselves.  The  movement  begins  slowly,  and  so 
continues  till  the  elastic  limit  of  the  metal  about  the  rivet-holes 
is  passed,  and  general  flow  takes  place  over  the  entire  cross-sec- 
tion, and  rupture  is  reached.  These  stages  in  the  test  of  a  joint 
are  well  defined,  except  when  the  plates  are  in  a  warped  condi- 
tion initially,  when  abnormal  micrometer  readings  are  observed. 

The  difference  in  behavior  of  a  joint  and  the  solid  metal 
suggests  the  propriety  of  arranging  tension  joints  in  boiler  con- 
struction and  elsewhere  as  nearly  in  line  as  practicable. 

The  efficiencies  of  the  joints  are  computed  on  the  basis  of 
the  tensile  strength  of  the  lengthwise  strips,  this  being  the 
direction  in  which  the  metal  of  the  joints  is  strained.  The 
efficiencies  here  found  are  undoubtedly  lowered  somewhat  by 
the  contraction  in  width  of  the  specimens,  causing  in  most  cases 
fractures  to  begin  at  the  edges  and  extend  towards  the  middle 
of  the  joint.  Of  the  entire  series,  88  joints  have  been  tested  ; 
the  J",  £",  and  £"  plates  yet  remain." 


576 


APPLIED  MECHANICS. 


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RIVET  METAL  FOR  RIVETED  JOINTS. 


579 


ch  Sections. 

iiiHiiiii 

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rt 
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—    0»          «                 COCO'^-'J-VOVOOOOO                 OO 

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Su 

^^            H^HO^^^-W^^            „£„£ 

£•= 

Q  « 

H             M                    M             M 

& 

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580 


APPLIED  MECHANICS. 


TABULATION   OF 

i"   STEEL 


No.  of 
Test. 


Sheet  Letters. 


Plate. 


Covers. 


Style  of  Joint. 


Nominal  Thick- 


Plate. 


Covers. 


Size  of 
Rivets 

and 
Holes. 


Actual 
Thick- 
ness of 
Plate. 


X3o8 
1309 


1310 
13" 


I3« 
IS'S 


1316 


1318 
1319 


1320 
1321 


1322 
1323 


1324 
1325 


inch. 

i 
i 


inch. 


ft 


inch. 

ft  and  | 
ft  and  i 


ft  and 
ft  and 


ft  and  I 
ft  and 


Hand* 
Hand* 


Hand* 
Hand* 


Hand* 


Hand* 
Hand* 


Hand* 
Hand* 


Hand! 
if  and  | 


inch. 


.242 
.242 


.242 

.249 


.244 
•243 

.248 
.242 


.244 
.246 


•245 
.240 


.243 
•243 


•245 
.247 


.248 
•  245 


TABULATION1  OF  RIVETED  JOINTS. 


581 


RIVETED  JOINTS. 
PLATE. 


Sectional  Area 
of  Plate. 

Searing 
Surface 
of 
Rivets. 

Shear- 

.  ing   . 
Area  of 
Rivets. 

Tensile 

Strength 
of  Plate, 
pounds 
per 
Square 
Inch. 

Maximum  Stress  on  Joint  per  Sq.  In. 

Effici- 
ency of 
Joint. 

Tension 
on  Gross 
Section  of 
Plate. 

Tension 
on  Net 
Section  of 
Plate. 

Compres- 
sion on 
Bearing 
Surface  of 
Rivets. 

Shear- 
ing on 
Rivets. 

Gross. 

Net. 

sq.  in. 

sq.  in. 

sq.  in. 

sq.  in. 

Ibs. 

Ibs. 

Ibs. 

Ibs. 

per  ct. 

2.360 

1.452 

.908 

3.682 

61740 

41690 

67770 

108370 

26720 

67-5 

2.360 

1.452 

.908 

3.682 



42180 

68560 

109640 

27040 

68.3 

2.541 

1.634 

.907 

3.682 

61740 

#2540 

66160 

119180 

29360 

68.9 

2    6lR 

1.681 

.034. 

3.682 

A.1  1  7O 

67160 

i  i  0810 

30660 

60    Q 

•*  •UXD 

3.745 

1.830 

•yj^r 

.915 

O"  wu* 

3.682 

61740 

i  *TO     i^9 

44920 

67380 

134750 

3349° 

uy  *y 

72.8 

2.739 

1.827 

.912 

3.682 



44520 

66750 

133720 

33"0 

72.1 

3.602 

1.486 

1.116 

5-300 

61740 

40700 

71270 

94890 

19980 

65.9 

2.541 

«.452 

1.089 

5.300 

40000 

70000 

93330 

19180 

64.8 

3.750 

1.652 

1.098 

5.300 

61740 

40980 

68210 

102630 

21260 

66.4 

3.773 

1.665 

1.107 

5-300 



UHSO 

68980 

103750 

21670 

67.1 

2.942 

1,840 

1.  102 

5-300 

61740 

42180 

67450 

112610 

23420 

68.3 

2.882 

1.802 

I.oSo 

5.300 

62660 

43000 

68770 

"4750 

23380 

68.6 

3.100 

2.007 

1-093 

5-300 

O2OOO 

43060 

66520 

122140 

25190 

68.7 

3.100 

2.007 

1.093 

5.300 



44030 

68000 

124880 

25750 

70.3 

3.310 

2.207 

I.I03 

5.300 

59180 

43040 

64540 

129150 

26880 

72.7 

3.337 

2.225 

1.  112 

5.300 



438lo 

65700 

131460 

27580 

74.0 

2.792 

1.490 

1.302 

7.216 

61740 

38650 

72480 

82870 

14950 

62.0 

2.756 

1.470 

1.286 

7.216 

39430 

73930 

84510 

15010 

63.8 

582 


APPLIED  MECHANICS. 


TABULATION   OF 
\"  STEEL. 


No.  of 
Test. 

Sheet  Letters. 

Style  of  Joint. 

Nominal  Thick- 
ness. 

Size  of 
Rivets 
and 
Holes. 

Actual 
Thick- 
ness of 
Plate. 

Plate. 

Covers. 

Plate. 

Covers. 

inch. 

inch. 

inch. 

inch. 

1326, 

G 

Q 

B 

B 
B 

p«^f 

i' 

t 

A 

if  and  J 
if  and  | 

.241 

1328 
1329 

G 
G 

B 
C 

C 
C 

i 

A 

A 

if  andf 
if  and| 

.241 
.242 

\                        1 

1330 

L 

C 

D 

V"  2*"piteh.—  It 

J 

& 

if  and| 

.248 

1331 

H 

C 

D 

|P  o  p'5y  ooj^ 

i 

A 

ifand  J 

.246 

r       Hf 

1332 

H 

D 

E 

\,  —  -«"piw?  4 

i 

re 

i|and| 

.248 

1333 

H 

D 

E 

LCI  ^  roe      -^1  ]* 

j 

T^ 

ii  and  i 

.248 

1334 

M 

E 

E 

t             A 

J 

A 

if  and| 

•243 

1335 

O 

IP  ®  re  w§      Oil, 

j 

A 

if  and  J 

•245 

1336 

H 

C 

C 

r"         •"? 

i 

A 

iiandj 

.238 

1337 

L 

C 

C 

L_.        13'.'l3      ,^>  ^ 

| 

A 

igand* 

.252 

1338 

G 

B 

B 

jl  ^-^jj-^.  ' 

| 

A 

if  and  i 

.238 

1339 

F 

B 

B 

E°fe£°S? 

j 

A 

if  and  i 

.248 

1340 

G 
G 

B 
C 

C 
C 

'               J 

i 

* 

if  and  i 
if  and  i 

.240 
.242 

W          °} 

1342 

L 

C 

D 

} 

ft 

if  and  i 

.250 

1343 

L 

D 

D 

(°°&9d°Jf 

j 

A 

if  and  i 

.250 

1344 

L 

D 

E 

i             ^ 

| 

A 

if  and  r 

.250 

J345 

H 

D 

E 

P-  ^  I"4'.'24        -^li 

& 

iV 

if  andr 

•247 

G     3 

1346 
1347 

I 

I 

E 

E 

E 
E 

pro2fe^hac| 

1 

A 

if  and  i 

•  251 
.252 

1                 I 

TABULATION  OF  RIVETED  JOINTS. 


583 


RIVETED   JOINTS—  Continued. 
PLATE— Continued. 


Sectional  Area 
of  Plate. 

Bearing 
Surface 
of 
Rivets. 

Shear- 
ing 
Area  of 
Rivets. 

Tensile 
Strength 
of  Plate, 
pounds 
per 
Square 
Inch. 

Maximum  Stress  on  Joint  per  Sq.  In. 

Effici- 
ency of 
Joint. 

Tension 
on  Gross 
Section  of 
Plate. 

Tension 
on  Net 
Section  of 
Plate. 

Compres- 
sion on 
Bearing 
Surface 
of  Rivets. 

Shear- 
ing on 
Rivets. 

Gross. 

f 

Net. 

sq.  in. 

sq.  in. 

sq.  in. 

sq.  in. 

Ibs. 

Ibs. 

Ibs. 

Ibs. 

per  ct. 

2.892 

1.627 

1.265 

7.216 

62660 

40340 

71700 

92210 

16170 

64.4 

2.909 

1.638 

1.271 

7.216 

41280 

73310 

94480 

16640 

65-9 

3-075 

1.810 

1.265 

7.216 

62660 

42290 

71850 

102810 

18020 

67-5 

3.088 

i.  817 

1.271 

7.216 



42750 

72650 

103860 

18290 

68.2 

3.348 

2.046 

1.302 

7.216 

61470 

43100 

70530 

110830 

20000 

70.1 

3-321 

2.029 

1.291 

7.216 

59180 

4145° 

67840 

106620 

19080 

70.0 

3-534 

2.232 

1.302 

7.216 

59180 

41820 

66210 

113500 

20480 

70.7 

3-534 

2.232 

1.302 

7.216 

42760 

67710 

116070 

20940 

72-3 

3.645 

2.369 

1.276 

7.216 

58170 

44650 

68700 

127550 

22550 

76.8 

3-675 

2.389 

1.286 

7.216 

64170 

43050 

66230 

123030 

21930 

67.1 

3-125 

2.084 

1.041 

6.013 

59180 

41310 

61960 

124020 

21470 

69.8 

3-309 

2.206 

i  .103 

6.013 

61470 

42000 

62990 

125990 

23IIO 

68.3 

2.856 

1.428 

1.428 

9-425 

62660 

40620 

81230 

81230 

I23IO 

64.8 

2.976 

1.488 

1.488 

9-425 

61740 

36290 

72580 

72580 

11460 

58.8 

3.060 

1.620 

1.440 

9-425 

62660 

38660 

73020 

82150 

I255o 

61.7 

3.088 

1.636 

1.452 

9-425 



38000 

71730 

80820 

12450 

60.6 

3.378 

1.878 

1.500 

9-425 

61470 

37800 

68000 

85130 

13550 

61.5 

3-375 

1-875 

1.500 

9-425 



39000 

70200 

87750 

13970 

63.4 

3.560 

2.060 

1-500 

9-425 

61470 

39130 

67640 

92870 

14780 

63.6 

3-52° 

2.038 

1.482 

9-425 

59180 

40450 

69860 

96070 

15110 

68.3 

3-765 

2.259 

1-506 

9-425 

60480 

43590 

72640 

108960 

17410 

72.1 

3.780 

2.268 

1.512 

9-425 



41420 

69030 

10354° 

16610 

68.5 

APPLIED  MECHANICS. 


TABULATION   OF 

\"  STEEL 


No.  of 
Test. 

Sheet  Letters. 

Style  of  Joint.      0 

Nominal  Thick- 
ness. 

Size  of 
Rivets 
and 
Holes. 

Actual 
Thick- 
ness of 
Plate. 

Plate. 

Covers. 

Plate. 

Covers. 

1348 
J349 

1350 
1351  • 

1352 
1353 

1354 
1355* 

H 
G 

N 
N 

H 
H 

I 
I 

C 
C 

D 
D 

D 
E 

E 
E 

C 
C 

D 
D 

D 
E 

E 
E 

inch. 
i 

i 

* 
i 

i 

i 

t 

i 

inch. 
& 

ft 

T3* 
& 

T3* 
A 

1% 
4 

inch. 
ft  and  i 
if  and  i 

H  and  i 
Hand  i 

ti  and  i 
H  and  i 

ft  and  i 
if  and  i 

inch. 
.244 
•  239 

.250 
.249 

.247 
.248 

.252 
.251 

r       ~r 

jp-^-qi 

v              JL 

EJE$ 

r       "fl 

™^4 

r        n 

f"  STEEL 


1356 

A 

I 

I 

\-mssri 

A  andf 

•365 

1357 

A 

I 

I 

?       O5        Q 

* 

i 

| 

T%  and  i 

•364 

1358 

A 

I 

J 

-ttfiar- 

i 

f 

} 

A  and  I 

•365 

»359 

A 

J 

J 

f  W  10'/49  ^  ~ 

i 

J 

&and| 

.366 

r 
i          i 

V   _  ^.    ...  ^J 

i 

1360 

A 

I 

J 

ooWoon! 

I 

% 

§ 

J 

Hand* 

.366 

1361 

A 

I 

I 

1 

«-     io!oO     ^| 

j 

1 

* 

IJandf 

•367 

i36* 

A 

J 

J 

EaSSSo" 

i 

t 

j 

Hand* 

.366 

1363 

A 

J 

J 

L_        11.25       ,_ 

i 

§ 

j 

H  and  J 

•365 

r 

V 

1364 

B 
B 

K 
K 

K 
K 

fo  OvXpb  c 

k-          12.00 

• 

! 

Hand* 
Hand* 

.388 
•390 

r~ 

*  Fractured  two  outside  sections  of  plate  at  each  edge  along  line  of  riveting;  the  two 
middle  sections  sheared  in  front  of  rivets. 


TABULATION  OF  RIVETED  JOINTS. 


585 


RIVETED   JOINTS— Continued. 
PLATE— Continued. 


Sectional  Area 
of  Plate. 

Bearing 
Surface 
of 
Rivets. 

Shear- 
ing 
Area  of 
Rivets. 

Tensile 
Strength 
of  Plate, 
pounds 
per 
Square 
Inch. 

Maximum  Stress  on  Joint  per  Sq.  In. 

Effici- 
ency of 
Joint. 

Tension 
on  Gross 
Section 
of  Plate. 

Tension 
on  Net 
Section 
of  Plate. 

Compres- 
sion on 
Bearing 
Surface 
of  Rivets. 

Shear- 
ing on 
Rivets. 

Gross. 

Net. 

sq.  in. 

sq.  in. 

sq.  in. 

sq.  in. 

Ibs. 

Ibs. 

Ibs. 

Ibs. 

per  ct. 

3.204 

1.984 

1.220 

7-854 

59180 

38700 

62490 

101620 

15790 

65.4 

3-131 

1.936 

t-195 

7-854 

62660 

42890 

69370 

112380 

17100 

68.4 

3.442 

2.192 

i  .250 

7-854 

55740 

42960 

67460 

118300 

18830 

77.1 

3.426 

2.181 

1-245 

7-854 

.   ... 

41780 

65640 

114980 

18230 

74-9 

3-554 

2.319 

1-235 

7-854 

59180 

435*° 

66690 

125220 

19690 

73-5 

3-569 

2.329 

1.240 

7-854 



43830 

67170 

126160 

19920 

74.1 

3.780 

2.520 

i  .260 

7-854 

60480 

44580 

66870 

133730 

21450 

73-7 

3-765 

2.510 

1-255 

7-854 



44410 

66610 

133230 

21290 

73-4 

PLATE. 


3-559 

3.190 

1.369 

3.682 

54260 

40460 

65740 

105170 

39100 

74-6 

3-549 

2.184 

1.365 

3.682 

39420 

64060 

102490 

38000 

72.6 

3.829 

2.460 

1.369 

3.682 

54260 

39780 

61910 

111250 

41360 

73-3 

3-843 

2.471 

1-372 

3-682 

39060 

60740 

109400 

40770 

72.0 

3-843 

2.  196 

1.647 

5-300 

54«6o 

37000 

64750 

86330 

26830 

68.2 

3-854 

2.202 

1.652 

5-300 



37050 

64840 

86430 

26940 

68.3 

4.118 

2.471 

1.647 

5.300 

54260 

37450 

62400 

93620 

29090 

69.0 

4.106 

2.464 

1.642 

5-300 

38040 

63390 

95130 

29470 

70.1 

4.656 

2.910 

1.746 

5-300 

59730 

41820 

66910 

111510 

36730 

70.0 

4.680 

2.925 

1-755 

5-300 

42000 

67200 

1  12000 

37090 

70-3 

586 


APPLIED  MECHANICS. 


TABULATION    OF 
|"   STEEL 


No.  of 
Test. 

Sheet  Letters. 

Style  of  Joint. 

Nominal  Thick- 
ness. 

Size  of 
Rivets 
and 
Holes. 

Actual 
Thick- 
ness cf 
Plate. 

Plate. 

Covers. 

Plate. 

Covers  . 

1366 
1367 

1368 
1369 

1370 

'371 

1372 
1373 

1374 
1375 

1376 
'377 

1378 
'379* 

1380 
1381 

1382 

1384 
1385 

C 
B 

A 
A 

B 
B 

B 
C 

D 

E 

D 
H 

D 
D 

B 
E 

B 
E 

F 
E 

K 
K 

J 

J 

K 

J 

L 
K 

H 

N 

L 
L 

M 
M 

J 
K 

K 
G 

H 

N 

K 
L 

J 
J 

K 

J 

L 
L 

N 
N 

M 
M 

M 
M 

K 
K 

K 
K 

H 

N 

E°83 

inch. 

1 
f 

* 

f 
f 

f 
f 

t 
t 

t 

i 

f 
f 

t 
1 

f 

f 

f 
f 

inch, 
i 

i 

inch. 

Hand* 
Hand* 

Hand| 

Hand| 
Hand* 

ttandf 
Handf 

Handf 
Hand* 

Hand  i 
Hand  i 

H  and  i 
H  and  i 

Hand  i 
Hand  i 

inch. 

•367 
•387 

•369 
.366 

•389 
.388 

•385 
•367 

.376 
.380 

•383 
•371 

•383 
•385 

.388 
.381 

.388 
.383 

.381 
.380 

t°SN 

j 

r~ 

L;,_,,__.h_._ 
pocwSbo 

f-            12.00            ~ 

r 

1 

jt 

*, 

f 

r~ 

lo  o  o/o  o  o 

|r      2^'pifcii;    "* 

r 

1 

V            J 

1 
i 

[qo2Soo-o; 

LJ 

r^»i  phcST"^ 

O  O  u/o  '&£>] 

f      J 

'O  O^O'CK) 

I 

r 

\ 

*  Test  discontinued  soon  after  passing  maximum  load, 
t  Test  discontinued  at  maximum  load. 


TABULATION  OF  RIVETED  JOINTS. 


587 


RIVETED   JOINTS—  Continued. 
PLATE— Continued. 


Sectional  Area 
of  Plate. 

Bearing 
Surface 
of 
Rivets. 

Shear- 
ing 
Area  of 
Rivets. 

Tensile 
Strength' 
of  Plate, 
pounds 
per 
Square 
Inch. 

Maximum  Stress  on  Joint  per  Sq.  In. 

Effici- 
ency of 
Joint. 

Tension 
on  Gross 
Section  of 
Plate. 

Tension 
on  Net 
Section  of 
Plate. 

Compres- 
sion on 
Bearing 
Surface  of 
Rivets. 

Shear- 
ing on 
Rivets. 

Gross. 

Net. 

sq.  in. 

sq.  in. 

sq.  in. 

sq.  in. 

Ibs. 

Ibs. 

Ibs. 

Ibs. 

per  ct. 

4-683 

3-031 

1.652 

5-3oo 

57870 

41040 

63410 

116340 

36260 

70.9 

4.938 

3-197 

1.741 

5-3oo 

59730 

40910 

63180 

116030 

38110 

68.5 

4-151 

2.214 

1-937 

T     O22 

7.216 
7.216 

54260 

35000 
34180 

65620 
64140 

75000 

j  20130 

T  f\  jfto 

64-5 

67.O 

4.668 

2.626 

•*-y-" 
2.042 

7.216 

59730 

o4lov 
36870 

65540 

73*5O 

84280 

*y4°° 

23850 

61.7 

A  grg 

2  .  6lQ 

2    O37 

7216 
•  *AU 

oftoAO 

69220 

8QOOO 

25120 

gr    2 

4»UDU 

4.916 

2-895 

~-wJ/ 
2.021 

7.216 

59730 

j°y4u 

38730 

65770 

oyvA-"-» 

94210 

26390 

I45*4 

64.8 

4.672 

2.745 

I.927 

7.216 

57870 

39010 

65660 

94580 

25260 

67.4 

5.076 

3.102 

1.974 

7.216 

53730 

37960 

62120 

97620 

26700 

7O.6 

5-T30 

3-135 

J-995 

7.216 

'  58340 

39810 

65140 

102360 

28300 

68.2 

5-450 

3-439 

2.  Oil 

7.216 

53730 

38920 

61670 

105470 

29390 

72.4 

5.290 

3-342 

1.948 

7.216 

56670 

39870 

63110 

108260 

29230 

70.4 

5-745 

3-734 

2.  Oil 

7.216 

53730 

40560 

62400 

115860 

32290 

75-5 

5T7C 

3     ire  A 

2.O2I 

7216 

AOTOO 

62620 

116280 

Q2CTO 

7C    "7 

•  7/D 

4-656 

•7D4 

2.328 

2.328 

tWAV 
9-425 

59730 

4.U/UV 

34500 

69010 

69010 

o*57u 

\  17050 

/»>•/ 

57-8 

4-572 

2.286 

2.286 

9-425 

58340 

33440 

66880 

66880 

16220 

57-3 

4-947 

2.619 

2.328 

9.425 

59730 

35590 

67230 

75640 

18680 

59-6 

4-883 

3-585 

2.298 

9-425 

58340 

34730 

65610 

73800 

17990 

59-5 

5-140 

2.854 

2.286 

9-425 

54290 

35490 

63930 

79810 

19360 

65-4 

5.126 

2.846 

2.280 

9.425 

58340 

35840 

64550 

80570 

19490 

61.4 

588 


APPLIED  MECHANICS. 


TABULATION    OF 

i"   STEEL 


Sheet  Letters. 

Style  of  Joint. 

Nominal  Thick- 
ness. 

Size  of 

Actual 

No.  of 

TpQt 

Rivets 

Thick- 

Holes. 

Plate. 

Plate. 

Covers. 

Plate. 

Covers. 

L                     JJ 

inch. 

inch. 

inch. 

inch. 

1386* 

H 

L 

L 

n  Pf^MrT^"  r-\ra 

t 

i 

Band  i 

.368 

1387 

G 

L 

M 

fr^L.     ra2r  ^  *2J£ 

f 

i 

Hand  i 

•365 

1388 
1389 

D 
D 

M 
M 

M 
M 

1 
f 

i 
| 

iiandi 
If  and  i 

•385 
.386 

"» 

1390* 

C 

G 

G 

t^gM^l 

t 

i 

if  and  i 

•372 

c 

H 

L 

K  ~*~B     -^1  3.  12^  J^-^l^ 

i 

is  an(j  j 

,60 

1392 

C 

L 

L 

M 

^      ,>       .A 

f 

i 

Hand  i 

•374 

b  JEffi  cS 

c 

L 

N 

\ 

?^-  ^  JftCi  ^* 

_ 

§ 

i 

Hand  i 

1394+ 

D 

I 

M 

( 

Sr^^ig^-  '«^.f  »  ^Jft 

t 

i 

Hand  i 

.386 

J39S 

D 

M 

M 

M  SS2  ^ir 

f 

I 

Hand  i 

•383 

f                   ^T 

*  Test  discontinued  after  passing  maximum  load, 
t  Test  discontinued  before  fracture  was  complete. 


TABULATION  OF  RIVETED  JOINTS. 


589 


RIVETED  JOINTS— Otfzww*/. 

PLATE— Continued. 


Sectional  Area 
of  Plate. 

B  earing 
Surface 
of 
Rivets. 

Shear- 

*  ing  f 
Area  of 

Rivets. 

Tensile 
Strength 
of  Plate, 
pounds 
per 
Square 
Inch. 

Maximum  Stress  on  Joint  per  Sq.  In. 

Effici- 
ency of 
Joint. 

Tension 
on  Gross 
Section  of 
Plate. 

Tension 
on  Net 
Section  of 
Plate. 

Compres- 
sion on 
Bearing 
Surface  of 
Rivets. 

Shear- 
ing on 
Rivets. 

Gross. 

Net. 

sq.  in. 

sq.  in. 

sq.  in. 

sq.  in. 

Ibs. 

Ibs. 

Ibs. 

Ibs. 

per  ct. 

5-244 

3.036 

2.208 

9-425 

56670 

37010 

63930 

87910 

20590 

65-3 

5-205 

3-°i5 

2.190 

9-425 

53840 

36750 

63450 

87350 

20300 

68.2 

5-775 

3-465 

2.310 

9-425 

53730 

37490 

62480 

937io 

22970 

69.8 

5-790 

3-474 

2.316 

9-425 



3736o 

62260 

93390 

21890 

69-5 

4.881 

3.021 

i.  860 

7.854 

57870 

39000 

63010 

102340 

24240 

67-4 

4.841 

2.996 

1-845 

7.854 



39520 

63850 

103690 

24360 

68.3 

5-143 

3-273 

1.870 

7-854 

57870 

39840 

62590 

109540 

26080 

68.9 

5-1" 

3-251 

i.  860 

7-854 



40420 

63550 

111070 

26310 

69.8 

5-555 

3-625 

1.930 

7.854 

53730 

39340 

60290 

113240 

27830 

73-2 

5-496 

3-S8« 

1-915 

7.854 

***** 

40300 

61850 

115660 

28200 

75-o 

5QO  APPLIED  MECHANICS. 


SINGLE-RIVETED   BUTT-JOINTS,  STEEL   PLATE. 
DESCRIPTION    OF   TESTS   AND    DISCUSSION    OF   RESULTS. 

"  The  following  tests  complete  a  series  of  two  hundred  and 
sixteen  single-riveted  butt-joints  in  steel  plates,  in  which  the 
thickness  of  the  plates  ranged  from  J"  to  J",  and  the  size  of 
the  rivets  from  TV'  to  iTyx  diameter. 

The  plates  were  annealed  after  shearing  to  size,  the  edges 
opposite  the  joint  milled  to  the  finished  width ;  the  holes  were 
drilled  and  rivets  machine-driven.  Iron  rivets  were  used 
throughout,  except  in  some  of  the  f "  joints. 

Tensile  tests  of  the  plates  and  rivet-metal,  together  with 
the  tests  of  the  joints  in  £"  and  fx/  plate,  are  contained  in  the 
Report  of  Tests  of  1885,  Senate  Document  No.  36,  Forty-ninth 
Congress,  first  session. 

The  tests  herewith  presented  comprise  the  details  and  tab- 
ulation of  joints  in  J-",  f",  and  f"  thickness  of  plate,  a  portion 
of  which  were  tested  hot. 

The  gauged  length  in  which  elongations  and  sets  were 
measured  was  5";  2,\"  each  side  of  the  centre  line  of  the  joint. 

During  the  progress  of  testing  the  same  characteristics  were 
displayed  which  were  referred  to  in  the  previous  report.  The 
joints  were  very  rigid  under  the  early  loads.  This  rigidity  is 
overcome  by  loads  which  exceed  the  friction  between  the  plate 
and  covers,  after  which  the  stretching  proceeded  slowly  with 
some  fluctuations  till  elongation  of  the  metal  of  the  net  section 
became  general ;  the  metal  under  compression  in  front  of  the 
rivets  yielding,  also  the  rivets  themselves. 

The  behavior  of  joints  in  different  thicknesses  of  plate  is 
substantially  the  same,  and  an  examination  of  the  results  shows 
that  when  exposed  to  similar  conditions  the  strength  per  unit 


SINGLE-RIVETED  BUTT-JOINTS,    STEEL  PLATE.        591 

of  fractured  metal  is  nearly  the  same,  whether  J"  or  f "  plate  is 
used. 

It  will  not  be  understood  from  this,  however,  that  as  a  con- 
sequence the  same  efficiency  may  be  obtained  in  different 
thicknesses  of  plate  for  single-riveted  work,  because  it  will  be 
seen  that  certain  essential  conditions  change  as  we  approach 
the  stronger  joints  in  different  thicknesses  of  plate. 

A  riveted  joint  of  the  maximum  efficiency  should  fracture 
the  plate  along  the  line  of  riveting,  for  it  is  clear  that  if  failure 
occurs  in  any  other  manner,  as  by  shearing  the  rivets  or  tear, 
ing  out  the  plate  in  front  of  the  rivet-holes,  there  remains  an 
excess  of  strength  along  the  line  of  riveting,  or  in  other  words 
along  the  net  section  of  metal — if  in  a  single-riveted  joint-^ 
which  has  not  been  made  use  of ;  but  when  fracture  occur* 
along  the  net  section  an  excess  of  strength  in  other  directions-,- 
is  immaterial. 

If  the  strength  per  unit  of  metal  of  the  net  section  was  con. 
stant,  it  would  be  a  very  simple  matter  to  compute  the  effi- 
ciency of  any  joint,  as  it  would  merely  be  the  ratio  of  the  net 
to  the*gross*areas  of  the  plates. 

The  tenacity  of  the  net  section,  however,  varies,  and  this 
variation  extends  over  wide  limits. 

In  the  present  series  there  is  an  excess  in  strength  of  the 
net  section  over  the  strength  of  the  tensile  test-pieces  in  all 
joints. 

Special  tables  have  been  prepared  showing  this  behavior. 

The  efficiencies  shown  in  Table  No.  I  are  obtained  by  divid- 
ing the  tensile  stress  on  the  gross  area  of  plate  by  the  tensile 
strength  of  the  plate  as  represented  by  the  strength  of  the  ten- 
sile test-strip,  stating  the  values  in  per  cent  of  the  latter. 

Table  No.  2  exhibits  the  differences  between  the  efficien- 
cies of  the  joints  and  the  ratios  of  net  to  .gross  areas  of  plate. 
If  the  tenacity  of  net  section  remained  constant  per  unit  of 


APPLIED  MECHANICS. 


area,  the  efficiencies  in  Table  No.  I  would,  as  above  explained^ 
be  identical  with  the  ratios  of  net  to  gross  areas  of  plate,  and 
the  values  in  this  table  reduced  to  zero. 

Table  No.  3  shows  the  excess  in  strength  of  the  net  section 
of  the  joint  over  the  strength  of  the  tensile  test-strip  in  per 
cent  of  the  latter. 

Table  No.  4  exhibits  the  compression  on  the  bearing-surface 
of  the  rivets  in  connection  with  the  excess  in  tensile  strength 
of  the  net  section  of  plate. 

Table  No.  I  is  valuable  in  showing  at  once  the  value  of 
different  joints  wherein  the  pitch  of  the  rivets  and  their  diame- 
ters vary. 

It  is  seen  there  is  considerable  latitude  allowed  in  the  choice 
of  rivets  and  pitch  without  materially  changing  the  efficiency 
of  the  joint ;  thus  in  J"  plate, 

|"  rivets  (driven),  if"  pitch,  72.4  per  cent  efficiency, 
|"  rivets  (driven),  2j"  pitch,  73.3  per  cent  efficiency, 
•J"  rivets  (driven),  2§"  pitch,  71.5  per  cent  efficiency, 
i"  rivets  (driven),  2^"  pitch,  70.3  per  cent  efficiency, 
i"  rivets  (driven),  2J-"  pitch,  73.8  per  cent  efficiency, 

give  nearly  the  same  results. 

In  these  examples  the  ratios  of  net  to  gross  areas  of  plate 
range  from  60  to  67  per  cent,  while  the  rivet-areas  range  from 
.3067  square  inch  to  .7854  square  inch.  The  actual  areas  of 
net  sections  of  plate  and  rivets  are  as  follows : 


f"  rivets. 

f  "  rivets. 

I"  rivets. 

i"  rivets. 

sq.  in. 
.  3067 

sq.  in. 
.4418 

sq.  in. 
.6013 

sq.  in. 

•7854 

Plate  

1.486 

2.2O7 

2.232 

j  2.259 

(  2.319 

SINGLE-RIVETED  BUTT-JOINTS,    STEEL  PLATE.        593 


The  areas  of  the  rivets  stand  to  each  other  as  the  following 
numbers : 

196  256 


100 


144 


and  the  net  areas  of  the  plate  to  each  other  as 
100       149       150 


152 
156 


From  these  illustrations  it  appears  that  to  attain  the  same 
degree  of  efficiency  in  this  quality  of  metal,  although  that 
efficiency  is  probably  not  the  highest  attainable,  a  fixed  ratio 
between  rivet  metal  and  net  section  of  plate  is  not  essential. 

In  J"  plate  with  -J"  rivets  the  efficiencies  of  the  joints  tested 
cold  are  nearly  constant  over  the  range  of  pitches  tested. 

The  efficiencies  and  the  ratio  of  net  to  gross  areas  of  plate 
are  as  follows : 


Pitch. 

ii" 

2" 

•4" 

2j" 

per  cent. 

per  cent. 

per  cent. 

per  cent. 

Efficiency.     .     .     . 

64-5 

66.3 

66.3 

66.4 

Ratio  of  areas  .     . 

53-4 

56.3 

58.9 

6l.I 

In  this  we  have  illustrated  a  case  which,  m  passing  from 
the  widest  pitch,  having  61.1  per  cent  of  the  solid  plate  left,  to 
the  narrowest  pitch,  which  had  53.4  per  cent  of  the  solid  plate, 
the  gain  or  excess  in  strength  in  the  net  section  almost  exactly 
compensated  for  the  loss  of  metal. 

In  Table  No.  3  the  average  of  all  the  joints  shows  the  high- 
est per  cent  of  excess  of  strength  in  the  narrowest  pitch,  and  a 
tendency  to  lose  this  excess  as  the  pitch  increases. 

Tests  of  detached  grooved  specimens  show  the  same  kind 


594 


APPLIED  MECHANICS. 


of  behavior,  but  as  they  are  not  subject  to  all  the  conditions 
found  in  a  joint,  the  analogy  does  not  extend  very  far. 

The  maximum  gain  in  strength  on  the  net  section,  not  for 
the  time  being  regarding  the  hot  joints,  and  disregarding  the 
exceptionally  high  value  of  joint  No.  1339,  J-"  plate,  was  21.2 
per  cent,  the  minimum  value  2.5  per  cent  of  the  tensile  test- 
strip.  In  other  forms  of  joints,  and  with  punched  holes  in  both 
iron  and  steel  plate,  illustrations  are  numerous  in  which  there 
have  been  large  deficiencies,  the  metal  of  the  net  section  fall- 
ing far  below  the  strength  of  the  plate. 

It  is  believed  to  have  been  amply  shown  that  increasing 
the  net  width  diminishes  the  apparent  tenacity  of  the  plate, 
although  other  influences  may  tend  to  counteract  this  tendency 
in  some  joints. 

In  order  to  compare  the  excess  in  strength  of  one  thickness 
of  plate  with  another  having  the  same  net  widths,  we  have  the 
following  table,  rejecting  those  joints  that  failed  otherwise 
than  along  the  line  of  riveting  in  making  these  averages :— 


Thickness  of  Plate. 

Width  of  plate  between  rivet-holes. 

i" 

H" 

i}" 

if" 

Ii" 

if" 

if" 

•*" 

3" 

V  
$"  

P.  ct. 

16.7 
18.4 
16.7 
17.7 

P.  ct. 

12.6 

13-7 
14-3 
16.3 

P.  ct. 

11.4 
12.7 

9-3 
14.2 

P.  ct. 

12.0 
13-5 
10.7 
14-5 

P.  Ct. 

13-4 

14.6 

9.1 

14.6 

7.6 

P.  ct. 

8.9 
12.9 
8.8 
12.7 
ii.  8 

P.  ct. 

"•5 
9.0 

8.2 

9-9 

IO.O 

P.  ct. 
13-1 
13-6 

12.2 

9.8 

IO.I 

P.  Ct. 

10.6 

*"  
f"  

3-5 

Average    of   all    thick- 
nesses     

16.2 

14.4 

12.3 

I2.Q 

11.9 

II.  0 

9-7 

ii.  8 

7.0 

The  excess  in  strength  is  generally  well  maintained  in  each 
of  the  several  thicknesses,  and  were  it  possible  to  retain  the 
same  ratio  of  net  to  gross  areas  of  plate,  and  at  the  same  time 


SINGLE-RIVETED  BUTT-JOINTS,    STEEL  PLATE.        595 

equal  net  widths  between  rivets,  it  would  seem  from  this  point 
of  view  feasible  to  obtain  the  same  degree  of  efficiency  in  thick 
as  in  thin  plates. 

The  following  causes,  however,  tend  to  prevent  such  a  con- 
summation. 

For  equal  net  widths  thick  plates  require  larger  rivets  to 
avoid  shearing  than  thin  ones,  the  diameters  of  the  rivets  being 
somewhat  increased  for  this  cause,  and  again  because  it  has 
become  necessary  to  increase  the  metal  of  the  net  section  in 
order  to  retain  a  suitable  ratio  of  net  to  gross  areas  of  plate. 

There  results  from  these  considerations  such  an  increase  in 
net  width  of  plate  that  the  excess  in  strength  displayed  by 
narrower  sections  is  lost,  and  consequently  the  result  is  a  joint 
of  lower  efficiency. 

The  data  relating  to  the  influence  of  compression  on  the 
bearing-surface  of  the  rivets,  on  the  tensile  strength  of  the 
plate,  as  shown  by  Table  No.  4  are  more  or  less  conflicting, 
However,  in  the  J"  plate,  in  which  the  most  intense  pressures 
are  found,  there  is  seen  a  pronounced  increase  in  tensile  strength 
as  the  pressures  diminish  in  intensity. 

It  is  probable  that  the  effects  of  intense  compression  would 
be  more  conspicuous  in  a  less  ductile  metal,  or  one  in  which 
the  ductility  had  been  impaired  by  punched  holes  or  otherwise. 

A  number  of  joints  were  tested  at  temperatures  ranging 
between  200°  and  700°  Fahr. 

The  heating  was  done  after  the  joints  were  in  position  for 
testing,  by  means  of  Bunsen  burners,  arranged  in  a  row  par- 
allel to  and  under  the  line  of  riveting. 

The  temperature  was  determined  with  a  mercurial  ther- 
mometer, the  bulb  of  which  was  immersed  in  a  bath  of  oil, 
contained  in  a  pocket  drilled  in  the  middle  rivet  of  the  joint. 

When  at  the  required  temperature  the  thermometer  was 
removed  from  the  joint,  a  dowel  was  driven  into  the  pocket  to 


596  APPLIED  MECHANICS. 

compensate  for  the  metal  of  the  rivet  which  had  been  removed 
by  the  drill,  and  then  loads  applied  and  gradually  increased  up 
to  the  time  of  rupture. 

Three  joints,  Nos.  1423,  1426,  and  1430,  were  tested  with- 
out dowels  in  the  oil-pockets. 

The  method  of  heating  was  to  raise  the  temperature  of  the 
joint,  as  shown  by  the  thermometer,  a  few  degrees  above  the 
temperature  at  which  the  test  was  made,  shut  off  the  gas- 
burners,  and  allow  the  temperature  to  fall  to  the  required  limit. 
The  temperature  fell  slowly,  draughts  of  cold  air  being  excluded 
from  the  under  side  of  the  joint  by  the  hood  which  covered 
the  gas-burners ;  the  upper  side  and  edges  of  the  joint  were 
covered  with  fine  dry  coal-ashes. 

The  results  show  an  increase  in  tensile  strength  when  heated 
over  the  duplicate  cold  joints  at  each  temperature  except  200° 
Fahr. 

From  200°  there  was  a  gain  in  strength  up  to  300°,  when 
the  resistance  fell  off  some  at  350°,  increased  again  at  400°,  and 
reached  the  maximum  effect  observed  at  500°  Fahr. ;  from  this 
point  the  strength  fell  rapidly  at  600°  and  700°. 

In  per  cent  of  the  cold  joint  there  was  a  loss  at  200°  of  3.2 
per  cent,  the  average  of  three  joints  ;  at  500°  the  gain  was  22.6 
per  cent,  the  average  of  four  joints.  The  maximum  and  mini- 
mum joints  at  this  temperature  showed  gains  of  27.6  per  cent, 
and  18.3  per  cent,  respectively. 

The  highest  tensile  strength  on  the  net  section  of  plate  was 
found  in  joint  No.  1433,  tested  at  500°  Fahr.,  where  81050 
pounds  per  square  inch  was  reached  against  a  strength  of  58000 
pounds  per  square  inch  in  the  cold  tensile  test-strip. 

The  hot  joints  showed  less  ductility  than  the  cold  ones, 
those  tested  at  200°  Fahr.  not  being  exempt  from  this  behav- 
ior, although  there  was  no  near  approach  to  brittleness  in  any. 

Three   joints,    Nos.    1418,   1420,  and    1424,   were   heated; 


SINGLE-RIVETED  BUTT-JOINTS,    STEEL  PLATE.        597 

strained  when  hot  with  loads  exceeding  the  ultimate  strength 
of  their  duplicate  cold  joints ;  the  loads  were  released,  and 
after  having  cooled  to  the  temperature  of  the  testing-room 
(No.  1424  cooled  to  150°  Fahr.)  were  tested  to  rupture,  and 
were  found  to  have  retained  substantially  the  strength  due 
their  temperature  when  hot. 

In  order  to  ascertain  that  the  time  intervening  between  hot 
straining  and  final  rupture  did  not  contribute  towards  the  ele- 
vation in  strength,  joint  No.  1434  was  strained  in  a  similar 
manner  with  a  load  approaching  rupture,  after  which  a  period 
of  rest  was  allowed  and  then  ruptured  without  material  gain 
in  strength. 

A  peculiarity  of  the  joints  fractured  at  400°  and  higher  tem- 
peratures was  the  comparatively  smooth  surface  of  the  frac- 
tured sections,  and  which  took  place  in  planes  making  angles 
of  about  50°  with  the  rolled  surface  of  the  plate. 

The  shearing-strength  of  the  iron  rivets  was  also  increased 
by  an  elevation  of  temperature. 

The  rivets  in  joint  No.  1410  at  the  temperature  of  350° 
sheared  at  43060  pounds  per  square  inch,  while  in  the  dupli- 
cate cold  joint  No.  1411  they  sheared  at  38530  pounds  per 
square  inch,  and  the  rivets  in  joint  No.  1398  at  300°  Fahr. 
were  loaded  with  46820  pounds  per  square  inch^and  did  not 
shear. 

Other  examples,  where  some  of  the  rivets  sheared  and  the 
plate  fractured  in  part,  showed  corresponding  gains  in  shearing- 
strength. 

The  almost  entire  absence  of  granular  fractures  in  these 
tests  is  a  feature  too  important  to  pass  by  without  special  men- 
tion." 


598. 


APPLIED  MECHANICS. 


TABULATION   OF   SINGLE- 
STEEL  PLATE. 


No.  of 
Test. 


Sheet  Letters. 


Plate. 


Covers. 


Style  of  Joint. 


Nominal  Thick 
ness. 


Plate.    Covers 


Size  of 
Rivets 

and 
Holes. 


Actual 
Thick- 
ness of 
Plate. 


1396 
1397 


1399 


1400 
1401 


1402 
1403 


1404 
1405 


1406 
1407 


1408 
1409 


1410 
1411 


m         w 


inch. 

* 

i 


inch. 


inch. 
Hand* 


Hand* 


Hand* 
H  and  f 


Hand 
Hand! 


Hand! 
Hand 


Hand 
Hand 


Hand  | 


Hand 
Hand 


inch. 

.481 
.484 


.484 
•483 


.486 
•  483 


.481 
.486 


.486 
.487 


.470 
.471 


.486 
.482 


.481 
.485 


TABULATION  OF  RIVETED  JOINTS. 


599 


RIVETED    BUTT-JOINTS. 


STEEL    PLATE. 


Sectional 
Area 
of  Plate. 

Bearing 
Surface 
of 
Rivets. 

Shear- 
ing 
Area  of 
Rivets. 

Tensile 
Strength 
of  Plate 
per 
Square 
Inch. 

Maximum  Stress  on  Joint  per  Sq.  In. 

i 

>> 

8« 

.2.5 

1* 
H 

Temperature  of  1 
Joint  in  Degrees 
Fahrenheit.  J 

Tension 
on  Gross 
Section  of 
Plate. 

Tension 
on  Net 
Section  of 
Plate. 

Compres- 
sion on 
Bearing 
Surface  of 
Rivets. 

Shear- 
ing on 
Rivets. 

Gross. 

Net. 

sq.in. 

sq.in. 

sq.  in. 

sq.  in. 

Ibs. 

Ibs. 

Ibs. 

Ibs. 

Ibs. 

per  ct. 

5-oS1 

2.886 

2.165 

5-3°i 

57180 

37750 

66070 

88080 

35970 

66.1 

200 

5.082 

2.904 

2.178 

5-301 

57180 

38980 

68220 

90960 

37370 

68.1 

5-445 

3-267 

2.178 

5-301 

57180 

4556o 

75960 

113960 

46820 

79-6 

300 

5-439 

3.266 

2.173 

5-301 

57180 

39260 

65400 

98290 

40290 

68.6 

5.842 

3-655 

2.187 

5-3oi 

57180 

37000 

59140 

98830 

40770 

64-7 

5-796 

3.622 

2.174 

5-3oi 

57180 

39420 

63080 

105100 

43100 

68.9 

350 

5.416 

2.891 

2-525 

7.216 

57180 

36890 

69110 

79i3o 

27690 

64-5 

5-477 

2.926 

5-55i 

7.216 

57180 

38250 

71600 

81730 

29030 

66.9 

250 

5.832 

3.281 

2.551 

7.216 

57180 

37910 

67380 

86670 

30640 

66.3 

5-854 

3-297 

2-557 

7.216 

57180 

43730 

77650 

IOO12O 

3548o 

76.4 

300 

5-997 

3-529 

2.468 

7.216 

59050 

44790 

76110 

108830 

37220 

76.0 

400 

6.010 

3-537 

2-473 

7.216 

59050 

39210 

66630 

953^0 

32660 

66.3 

6.561 

4.010 

2.551 

7.216 

60000 

39850 

65210 

102500 

36230 

66.4 

6.512 

3.982 

2.530 

7.216 

60000 

46610 

76220 

119980 

42060 

77-6 

500 

6.859 

4-334 

2-525 

7.216 

60000 

453oo 

71690 

123050 

43060 

75-5 

350 

6.916 

4.370 

2.546 

7.216 

60000 

40050 

63390 

I08800 

38530 

66.7 

... 

6oo 


APPLIED  MECHANICS. 


TABULATION   OF  SINGLE- 

STEEL  PLATE— Continued. 


No.  of 
Test. 


1412 
1413 


1414 


1416 

1417 


1418 
1419 


1420 
1421 


1422 
1423 


1424 
1425 


Sheet  Letters. 


Plate. 


Covers. 


Style  of  Joint. 


Nominal  Thick 
ness. 


Plate.    Covers 


inch. 


inch. 


Size  of 
Rivets 

and 
Holes. 


inch. 


H  and  i 
\\  and  i 


II  and  i 
Hand 


if  and  i 
Hand  i 


Hand 
Hand 


Hand! 
H  and  i 


H  and  i 
Hand  i 


Hand  i 
Hand  i 


Actual 
Thick- 
ness of 
Plate. 


inch. 


.484 
.481 


.472 

.468 


.468 
.482 


.481 

.482 


•479 
•483 


.469 
•473 


.484 
•483 


TABULATION  OF  RIVETED  JOINTS. 


601 


RIVETED    BUTT-JOINTS—  Continued. 

STEEL  PLATE— Continued. 


Sectional 
Area 
of  Plate. 

Bearing 
Surface 
of 
Rivets. 

Shear- 
•    ing 
Area  of 
Rivets. 

Tensile 
Strength 
of  Plate 
per 
Square 
Inch. 

Maximum  Stress  on  Joint  per  Sq.  In. 

Efficiency  of 
Joint. 

1  Temperature  of 
Joint  in  Degrees 
1  Fahrenheit. 

Tension 
on  Gross 
Section  of 
Plate. 

Tension 
on  Net 
Section  of 
Plate. 

Compres- 
sion on 
Bearing 
Surface 
of  Rivets. 

Shear- 
ing on 
Rivets. 

Gross. 

Net. 

sq.  in. 

sq.  in. 

sq.  in. 

sq.  in. 

Ibs. 

Ibs. 

Ibs. 

Ibs. 

Ibs. 

per  ct. 

5-813 
5-772 

2.909 
2.886 

2.904 
2.886 

9.425 
9-425 

57180^ 
57180 

35920 
34390 

71770 
68780 

71900 

68780 

92150 
21060 

62.8 
60.  i 

250 

6.023 

3-I91 

2.832 

9-425 

59050 

35000 

66020 

74430 

22360 

59-2 

5-967 

3-159 

2.808 

9-425 

59050 

40250 

76030 

85540 

25480 

68.1 

300 

6.327 

3-5I9 

2.808 

9-425 

59050 

34660 

62320 

78090 

23160 

60.3 

200 

6.512 

3.620 

2.892 

9-425 

60000 

36950 

66480 

83220 

25530 

61.5 

6.859 

3-973 

2.886 

9-425 

60000 

437io 

75460 

103880 

31810 

72.8 

(*) 

6.883 

3-991 

2.892 

9-425 

60000 

38720 

66770 

92150 

28270 

64-5 

... 

7-194 

4-320 

2.874 

9-425 

58000 

44840 

74670 

112250 

34230 

77-3 

(t) 

7-245 

4-347 

2.898 

9-425 

58000 

38740 

64570 

96850 

29780 

66.9 

6.167 

3.822 

2-345 

7.854 

59050 

39730 

64110 

104490 

31200 

67.2 

6.220 

3.855 

2.365 

7-854 

59050 

45420 

73250 

119450 

35840 

76.9 

400 

6.660 

4.240 

2.420 

7-854 

60000 

48950 

76890 

137110 

41510 

81.5 

<*) 

6.632 

4.317 

2-415 

7-854 

60000 

40600 

63860 

111490 

34280 

67.6 

*  Strained  while  at  the  temperature  of  400°  Fahr.  and  allowed  to  cool  before  rupture. 
+  Strained  while  at  the  temperature  of  500°  Fahr.  and  allowed  to  cool  before  rupture. 
t  Strained  while  at  the  temperature  of  500°  Fahr.,  then  cooled  to  150°  Fahr.  and  ruptured. 


6O2 


APPLIED  MECHANICS. 


TABULATION    OF    SINGLE- 
STEEL  PLATE— Continued. 


No.  of 
Tesc. 


Sheet  Letters. 


Plate. 


Covers. 


Style  of  Joint. 


Nominal  Thick 
ness. 


Plate. 


Covers 


Size  of 
Rivets 

and 
Holes. 


Actual 
Thick- 
ness of 
Plate. 


1426 
1427 


1428 
1429 


i43° 
I431 


I432 


1434 
1435 


1437 


1438 
*439 


inch. 


inch. 


inch. 


:  A  and 
T>B  and  i\ 


^  and 


ij^and 


IT'S  and 
1^5  and  14. 


i  A  and  it 


inch. 


•474 
•475 


•479 
.465 


.484 
•483 


.484 
.481 


.472 

•475 


.482 
.482 


.484 
•485 


TABULATION  OF  RIVETED  JOINTS. 


603 


RIVETED    BUTT-JOINTS—  Continued.     . 

STEEL  PLATE— Continued. 


Sectional 
Area  of 
Plate. 

Bearing 
Surface 
of 
Rivets. 

Shear- 
ing 
Area  of 
Rivets. 

Tensile 
Strength 
of  Plate 
per 
Square 
Inch. 

Maximum  Stress  on  Joint  per  Sq.  In. 

1  Efficiency  of 
Joint. 

Temperature  of 
Joint  in  Degrees 
Fahrenheit.  | 

Tension 
on  Gross 
Section  of 
Plate. 

Tension 
on  Net 
Section  of 
Plate. 

Compres- 
sion on 
Bearing 
Surface  of 
Rivets. 

Shear- 
ing on 
Rivets. 

Gross. 

Net. 

sq.in. 

sq.  in. 

sq.  in. 

sq.  in. 

Ibs. 

Ibs. 

Ibs. 

Ibs. 

Ibs. 

perct. 

6-053 

2.853 

3.200 

11.928 

5905° 

35070 

74410 

66340 

18630 

59-3 

300 

6.042 

2.836 

3.206 

11.928 

59050 

30420 

64810 

57330 

15410 

51-5 

6.471 

3-238 

3-233 

11.928 

60000 

40330 

80620 

80730 

21880 

67.2 

35o; 

6.278 

3-139 

3-139 

11.928 

59050 

33420 

66840 

66840 

17590 

56-5 

6.897 

3.620 

3-277 

11.928 

58000 

36390 

65150 

76590 

•  21040 

62.7 

700 

6.89? 

3-632 

3.260 

11.928 

58000 

33660 

63870 

71160 

19450 

58.0 

7.270 

4-043 

3.227 

11.928 

58000 

36140- 

65000 

81430 

22030 

62.3 

7-215 

3-968 

3-247 

11.928 

58000 

44490 

81050 

99040 

26960 

J76-7 

500 

6.193 

3.538 

2-655 

9.940 

59050 

36670 

64190 

85540 

22850 

62.0 

6.232 

3-560 

2.672 

9.940 

59050 

36200 

63870 

84420 

22690 

61.2 

... 

6.632 

3-921 

2.711 

9.940 

60000 

42430 

71610 

103580 

.  28250 

70.5 

500 

6.632 

3-921 

2.711 

9.940 

60000 

38720 

65440 

94730 

25840 

64.5 

... 

6.965 

4-243 

2.722 

9.940 

58000 

46630 

76550 

119320 

32680 

80.3 

500 

6.974 

4-246 

2.728 

9.940 

58000 

38900 

63890 

99400 

27290 

67.0 

604 


APPLIED  MECHANICS. 


TABULATION  OF  SINGLE- 
STEEL  PLATE— Continued. 


No.  of 
Test. 


Sheet  Letters. 


Plate.  Covers. 


Style  of  Joint. 


Nominal  Thick- 


Plate. 


Size  of 
Rivets 

and 
Holes. 


Actual 
Thick- 
ness of 
Plate. 


1440 
1441 


1443 


1444 
I44S 


1446 
1447 


1448 
1449 


1452 
MSB 


W 
W 


inch. 


inch. 


inch. 


^  and  i£ 
&  and 


A  and 


13  and  £ 
if  and  £ 


Hand  | 
Hand* 


Hand* 
if  and  i 


ifandl 


II  and  i 
Hand  i 


inch. 


.482 
•483 


•483 
.484 


.621 
.624 


.616 
.624 


.621 
.624 


.610 
.611 


.624 
.620 


TABULATION  OF  RIVETED  JOINTS. 


605 


RIVETED  BUTT-JOINTS—  Continued. 

STEEL  PLATE— Continued. 


Sectional 
Area 
of  Plate. 

Bearing 
Surface 
of 
Rivets. 

Shear- 
ing 
Area  of 
Rivets. 

Tensile 
Strength 
ol  Plate 
per 
Square 
Inch. 

Maximum  Stress  on  Joint  per  Sq.  In. 

*o 
>, 

8  . 

US 
£J° 
w 

Temperature  of 
Joint  in  Degrees 
Fahrenheit. 

Tension 
on  Gross 
Section 
of  Plate. 

Tension 
on  Net 
Section 
of  Plate. 

Compres- 
sion on 
Bearing 
Surface 
of  Rivets. 

Shear- 
ing on 
Rivets. 

Gross. 

Net. 

sq.  in. 

sq.  in. 

sq.  in. 

sq.  in. 

Ibs. 

Ibs. 

Ibs. 

Ibs. 

Ibs. 

perct. 

7.230 

4-519 

2.711 

9.940 

58000 

39180 

62690 

104500 

28500 

67-5 

200 

7.250 

4-528 

2.722 

9.940 

58000 

40630 

65060 

108220 

29630 

70.0 

7-564 

4-837 

2.717 

9.940 

57410 

38570 

60450 

107610 

29410 

67.2 

7-565 

4-843 

2.722 

9.940 

57410 

39160 

61170 

108830 

29800 

68.2 

6.986 

3.726 

3.260 

7.216 

55000 

33750 

63280 

72330 

32670 

60.  T 

7.020 

3-744 

3.276 

7.216 

55000 

34530 

64740 

74000 

33590 

62.7 

... 

7-392 

4.158 

3-234 

7.216 

55000 

36760 

65340 

84010 

37650 

66.0 

7.488 

4.212 

3.276 

7.216 

55000 

35120 

62440 

80280 

36440 

63.8 

... 

7.918 

4.658 

3.260 

7.216 

55000 

4193° 

71270 

101840 

46010 

76.2 

300 

7-956 

4.680 

3.276 

7.216 

55000 

36800 

62560 

89370 

40570 

66.9 

... 

8.241 

5-039 

3.202 

7.216 

57290 

39320 

64290 

101180 

44900 

68.6 

400 

8.249 

5.042 

3.207 

7.216 

57290 

36850 

60290 

94790 

42130 

64-3 

600 

7.488 

3-744 

3-744 

9-425 

55000 

32080 

64150 

64150 

25480 

58.3 

7.440 

3.720 

3.720 

9-425 

55000 

32060 

64110 

64110 

25300 

58-3 

6o6 


APPLIED   MECHANICS. 


TABULATION    OF  SINGLE- 
STEEL  PLATE- Continued, 


No.  of 
Test. 


Sheet  Letters. 


Plate. 


Covers. 


Style  of  Joint. 


Nominal  Thick- 


Plate. 


Size  of 
Rivets 

and 
Holes. 


Actual 
Thick- 
ness of 
Plate. 


T454 
H55 


1456 
M57 


1458 
1459 


1460 
1461 


1462 
1463 


1464 
1465 


1466 
1467 


inch. 


inch. 


inch. 


\\  and  i 
if  and  i 


\\ and  i 
\\  and  i 


\\  and  i 
\l  and  i 


Hand  i 
if  and  i 


\\ and  i 
H  and  i 


^8  and  \\ 


&  and  i 
^  and  i 


inch. 


.622 
.618 


.612 
.611 


.610 
.608 


.617 

.618 


.630 
.608 


.624 
.623 


.613 

.606 


TABULATION  OF  RIVETED  fOTNTS. 


607 


RIVETED    BUTT-JOINTS—  Continued. 

STEEL  PLATE— Continued. 


Sectional 
Area 
of  Plate. 

Bearing 
Surface 
of 
Rivets. 

Shear- 
ing 
Area  of 
Rivets. 

Tensile 
Strength 
of  Plate 
per 
Square 
Inch. 

Maximum  Stress  on  Joint  per  Sq.  In. 

Efficiency  of 
Joint. 

Temperature  of 
Joint  in  Degrees 
Fahrenheit. 

Tension 
on  Gross 
Section  of 
Plate. 

Tension 
on  Net 
Section  of 
Plate. 

Compres- 
sion on 
Bearing 
Surface 
of  Rivets. 

Shear- 
ing on 
Rivets. 

Gross. 

Net. 

sq.  in. 

sq.  in. 

sq.  in. 

sq.  in. 

Ibs. 

Ibs. 

Ibs. 

Ibs. 

Ibs. 

per  ct. 

7-93i 

4.199 

3-732 

9-425 

55000 

34120 

64440 

72510 

28710 

60.0 

... 

7.880 

4.172 

3.708 

9.425 

55000 

34000 

64220 

72250 

28420 

61.8 

.  .  . 

8.262 

4-590 

3.672 

9-425 

57290 

36490 

65680 

82110 

32000 

63.6 

8.249 

4-583 

3.666 

9-425 

57290 

36020 

64830 

81040 

31520 

62.8 

.  .  . 

8.662 

5.002 

3.660 

9-425 

57290 

37720 

65310 

89260 

38490 

65-8 

8.664 

5.016 

3-648 

9-425 

57290 

37540 

64850 

89170 

345io 

65-5 

... 

9-255 

5-553 

3.702 

9-425 

55940 

373oo 

62160 

93250 

36630 

66.6 

9.282 

5-574 

3.708 

9-425 

55940 

3700° 

61610 

92620 

36440 

66.1 

... 

8.259 

5.109 

3-r50 

7.854 

55000 

3578o 

57840 

93810 

37620 

65.0 

7.965 

4-925 

3-040 

7-854 

57290 

36960 

59770 

96840 

37500 

64.5 

... 

7-95° 

3-738 

4.212 

11.928 

55000 

31090 

66130 

58690 

20720 

56.5 

7-949 

3-744 

4-205 

11.928 

55000 

31090 

66020 

58780 

20720 

56.5 

.  .  . 

8.269 

4-131 

4.138 

11.928 

57290 

33*50 

66350 

66240 

22980 

57-8 

8.181 

4.090 

4.091 

11.928 

55940 

33240 

66250 

66240 

22720 

58.0 

6o8 


APPLIED  MECHANICS. 


TABULATION   OF   SINGLE- 
STEEL  PLATE-Co«//««^. 


No.of 
Test. 


Sheet  Letters. 


Plate. 


Covers. 


Style  of  Joint. 


Nominal  Thick- 
ness. 


Plate.    Covers. 


Size  of 
Rivets 

and 
Holes. 


1468 


1470 
1471 


1472 
M-73 


1476 
M77 


1478 


1480 
1481 


W 
W 


I 
N&' 


inch. 


inch. 


inch. 


^and 


T*Band 


^  and 


:^and 
^and 


TABULATION  OF  RIVETED  JOINTS. 


609 


RIVETED    BUTT-JOINTS— Continued. 

STEEL    PLATE— Continued. 


Sectional 
Area 
of  Plate. 

tearing 
Surface 
of 
Rivets. 

Shear- 
ing 
Area  of 
Rivets. 

Tensile 
Strength 
of  Plate 
per 
Square 
Inch. 

Maximum  Stress  on  Joint  per  Sq.  In. 

Efficiency  of 
Joint. 

I  Temperature  of 
Joint  in  Degrees 
1  Fahrenheit. 

Tension 
on  Gross 
Section  of 
Plate. 

Tension 
on  Net 
Section  of 
Plate. 

Compres- 
sion on 
Bearing 
Surface  of 
Rivets. 

Shear- 
ing on 
Rivets. 

Gross. 

Net. 

sq.in. 

q.  in. 

sq.  in. 

sq.  in. 

Ibs. 

Ibs. 

Ibs. 

Ibs. 

Ibs. 

per  ct. 

8.735 

4-597 

4.138 

11.928 

57290 

34260 

65110 

72330 

25090 

59-8 

8.690 

4-579 

4.111 

11.928 

57290 

34790 

66030 

73540 

25350 

60.7 

9.285 

5-107 

4.178 

11.928 

55940 

34980 

63600 

77740 

27230 

62.5 

9.240 

5.082 

4.158 

11.928 

55940 

3477° 

63220 

77270 

26930 

62.1 

8.239 

4.706 

3-533 

9.940 

55000 

36350 

63640 

84770 

30130 

66.1 

7.978 

4-552 

3-426 

9.940 

57290 

37100 

65020 

86400 

29780 

64.7 

8.362 

4-936 

3-426 

9.940 

57290 

38150 

64630 

93110 

32090 

66.5 

8.381 

4-950 

3-43i 

9.940 

57290 

38120 

64540 

93120 

32140 

66.5 

8.833 

5.402 

3-43i 

9.940 

57290 

38620 

63140 

99410 

343^0 

67-4 

8-739 

5-3I3 

3.426 

9.940 

57290 

38180 

62830 

97430 

3358o 

66.6 

9.240 

5-775 

3-465 

9.940 

55940 

38480 

61570 

102630 

35770 

68.7 

9-345 

5.841 

3-504 

9.940 

55940 

38410 

61430 

102440 

36110 

68.6 

... 

I 

7-8i3 

5.000 

2.813 

7-952 

55000 

37340 

58360 

103730 

36690 

67.9 

7-763 

4.968 

2-795 

7-952 

55000 

38440 

60060 

106760 

37520 

69-9 

6io 


APPLIED   MECHANICS. 


TABULATION   OF   SINGLE- 

STEEL   PLATE—  Continued. 

No.of 
Test. 

Sheet  Letters. 

Style  of  Joint. 

Nominal  Thick- 
ness. 

Size  of 
Rivets 
and 
Holes. 

Actual 
Thick- 
ness of 
Plate. 

Plate. 

Covers. 

Plate. 

Covers, 
inch. 

inch. 

inch. 

inch. 

1482 

Z 

K 

K 

i 

y7g 

1§  and  i 

•736 

1483 

Z 

K 

K 

t 

A 

Hand  i 

•757 

1484 

Z 

K 

K 

V 

i 

T7* 

11  and  i 

•742 

1485 

Z 

P 

P 

1 

i 

A 

If  and  i 

.762 

1486 
1487 

1488 
1489 

1490 

Z 
Z 

Z 
Z 

Z 

L 
L 

N 
O 

K 

M 
M 

N 

N 

i 
i 

i 
t 

A 

liana  i 
If  and  i 

If  and  i 
If  and  i 

1  1*5  and  i£ 

•749 
.764 

•745 
•735 

•723 

^ 

t          1 

1491 
1492 

Z 
Z 

P 

M 

P 
L 

\ 

OCK^/O^f" 

i 

i 

IT'S  and  i£ 
i  j'g  and  i^- 

•752 
.736 

bS^^ofi 

M93 

Z 

M 

M 

1 

pfiSSSzESIJ 

i 

A 

i*and.* 

•754 

1494 
M95 

Z 
Z 

N 
O 

O 
0 

( 

( 

i 

A 

ij^and  ij- 

.760 
.760 

"N 

TABULATION   OF  RIVETED  JOINTS. 


6-1 1 


RIVETED    BUTT-JOINTS—  Continued. 

STEEL  PLATE-G»»*/«»«*. 


Sectional 
Area 
of  Plate. 

Bearing 
Surface 
of 
Rivets. 

Shear- 
ing 
Area  of 
Rivets. 

Tensile 
Strength 
of  Plate 
per 
Square 
Inch. 

Maximum  Stress  on  Joint  per  Sq.  In. 

Efficiency  of 
Joint. 

Temperature  of 
Joint  in  Degrees 
Fahrenheit.  j 

Tension 
on  Gross 
Section  of 
Plate. 

Tension 
on  Net 
Section  of 
Plate. 

Compres- 
sion  on 
Bearing 
Surface 
of  Rivets. 

Shear- 
ing on 
Rivets. 

Gross. 

Net. 

sq.  in. 

sq.  in. 

sq.  in. 

sq.  in. 

Ibs. 

Ibs. 

Ibs. 

Ibs. 

Ibs. 

perct. 

8.847 

4-431 

4.416 

9-425 

59000 

31990 

63870 

64090 

30030 

54-2 

9.099 

4-557 

4-542 

9-425 

59000 

31980 

63860 

64070 

30870 

54-2 

9-475 

5.023 

4-452 

9-425 

59000 

34440 

64960 

73290 

34620 

58-3 

9-723 

5-151 

4-572 

9-425 

59000 

34700 

67340 

73790 

35800 

58.8 

IO.  112 

5.618 

4.494 

9.425 

59000 

35000 

63000 

78750 

37550 

59-3 

10.329 

5-745 

4-584 

0-425 

59000 

36780 

66130 

82870 

40310 

62.3 

10.624 

6.154 

4.470 

9-425 

59000 

38120 

65810 

90600 

42970 

64.6 

10.488 

6.078 

4.410 

9-425 

59000 

34000 

58670 

80860 

37830 

57-6 

9-233 

4-353 

4.880 

11.928 

59000 

31050 

65860 

58750 

24030 

52-6 

9^96 

4.520 

5-076 

11.928 

59000 

32000 

67940 

65340 

25740 

54-2 

10.179 

4-983 
5.089 

4.968 
5-090 

11.928 
i  i  .  928 

59°oo 
59000 

34270 
33770 

68430 
67540 

68640 
67520 

28590 
28810 

58.0 
57-2 

10.845 

5.7I5 

5-130 

ii  .928 

59000 

34900 

66230 

7378o 

31730 

10.838 

5-708 

5-I3° 

11.928 

59000 

35810 

67990 

75650 

32540 

60.6 

0?  THE 


7ERSIT7! 


6l2 


APPLIED   MECHANICS. 


TABULATION    OF    SINGLE- 
STEEL  PLATE—  Continued. 


No.  of 


1496 
M97 


1498 
1 499 


1500 
1501 


1502 
1503 


1504 


1506 


1508 


Sheet  Letters. 


Plate. 


Covers. 


O 
O  . 


Style  of  Joint. 


Nominal  Thick- 


Plate.  'Covers 


inch. 


inch. 


Size  of 
Rivets 

and 
Holes. 


inch. 


1 1*5  and  i£ 


^and 


i^  and 


1 3*5  and 


iT3B  and 
iT3sand 


iT35  and 


TBand 
T3B  and 


Actual 
Thick- 
ness of 
Plate. 


inch. 


•745 
.725 


•733 
•744 


762 
.727 


.722 
.741 


.722 
.762 


.727 
•735 


•737 
•753 


TABULATION  OF  RIVETED  JOINTS. 


613 


RIVETED    BUTT-JOINTS—  Continued. 

STEEL  PLANTS.— Continued. 


Sectional 
Area 
of  Plate. 

Bearing 

Surface 
of 
Rivets. 

Shear- 
ing 
Area  oj 
Rivets. 

Tensile 
Strength 
of  Plate 
per 
Square 
Inch. 

Maximum  Stress  on  Joint  per  Sq.  In. 

*o 
N 

'o.S 

££ 
w 

Temperature  of 
Joint  in  Degrees 
Fahrenheit. 

Tension 
on  Gross 
Section 
of  Plate. 

Tension 
on  Net 
Section 
of  Plate. 

Compres 
sion  on 
Bearing 
Surface 
of  Rivets 

Shear- 
ing on 
Rivets. 

Gross. 

Net. 

sq.  in. 

sq.  in. 

sq.  in. 

sq.  in. 

Ibs. 

Ibs. 

Ibs. 

Ibs. 

Ibs. 

per  ct. 

11.175 

6.146 

5-029 

11.928 

60420 

38470 

69940 

85480 

36040 

63.6 

10.890 

5.996 

4.894 

11.928 

60420 

37740 

68650 

83980 

34460 

62.4 

... 

9.624 

5-501 

4-123 

9.940 

59000 

35000 

61230 

81700 

33890 

57-o 

9.776 

5-572 

4.204 

10.030 

59000 

36470 

63990 

84810 

35550 

61.7 

... 

10.478 

6.192 

4.286 

9.940 

59000 

38760 

65590 

94750 

40850 

65-7 

10.004 

5.9I5 

4.089 

9.940 

59000 

36740 

62130 

89880 

36970 

62.2 

10.390 

6.329 

4.061 

9.940 

59000 

37930 

62270 

97050 

39650 

64-3 

10.663 

6-495 

4.168 

9.940 

59000 

40630 

65810 

90600 

42970 

68.8 

... 

9.761 

4.346 

5-415 

14.726 

59000 

29090 

65350 

52460 

19280 

49-3 

10.287 

4-572 

5.7IS 

14.726 

59000 

30010 

67520 

54010 

20960 

50.8 

... 

10.367 

4.914 

5-453 

14.726 

60420 

33610 

70900 

63890 

23660 

55-6 

10.474 

4.961 

5-513 

14.726 

60420 

33660 

71070 

63960 

23940 

55-7 

... 

11.070 

5-542 

5-528 

14.726 

60420 

3478o 

69470 

69650 

26140 

57-5 

11.310 

5.662 

5-648 

14.726 

60420 

34670 

69250 

69420 

26620 

57-4 

APPLIED  MECHANICS. 


TABULATION    OF    SINGLE- 
STEEL  PLATE— Continued. 


No.  of 
Test. 


Sheet  Letters. 


Plate. 


Covers. 


Style  of  Joint. 


Nominal  Thick 
ness. 


Plate. 


Size  of 
Rivets 

and 
Holes. 


Actual 
Thick- 
ness of 
Plate. 


1510 


1513 


'SIS 


1516 


1518 
1519 


1520 
1521 


1522 
1523 


inch. 


inch. 


inch. 


ft  and 


i ft  and 
iftand 


[ft  and  it 
[ft  and  it 


iftand  it 
iftand  it 


t A  and  it 
(ft  and  it 


iftand  it 
IT'S  and  it 


iftand  it 
iftand  it 


inch. 


•  748 
•755 


•750 
•764 


.760 
.746 


•749 
.741 


.756 
.741 


•763 
.718 


•742 
•754 


NOTE.— The  figures  in  larger  type  indicate  the  manner  of  failure.     When  different  kinds  ol 

than  one 


TABULATION  OF  RIVETED  JOINTS. 


6l5 


RIVETED    BUTT-JOINTS—  Continued. 

STEEL  PLATE—  Continued. 


Sectional 
Area  of 
Plate. 

B  earing 
Surface 
of 
Rivets. 

Shear- 
ing 
Area  of 
Rivets. 

Tensile 
Strength 
of  Plate 
per 
Square 
Inch. 

Maximum  Stress  on  Joint  per  Sq.  In. 

Efficiency  of 
Joint. 

Temperature  of 
Joint  in  Degrees 
Fahrenheit. 

Tension 
on  Gross 
Section  of 
Plate. 

Tension 
on  Net 
Section  of 
Plate. 

Compres- 
sion on 
Bearing 
Surface  of 
Rivets. 

Shear- 
ing on 
Rivets. 

Gross- 

Net. 

sq.  in. 

sq.  in. 

sq.  in. 

sq.  in. 

Ibs. 

Ibs. 

Ibs. 

Ibs. 

Ibs. 

per  ct. 

9.918 

5-243 

5-675 

12.272 

60420 

36120 

•68380 

76680 

29210 

59-7 

9.928 

5.209 

4.719 

12.272 

60420 

36940 

70400 

77710 

29880 

61.1 

10.328 

5.640 

4.688 

12.272 

59000 

33730 

61770 

74320 

28390 

57-0 

10.505 

5-73° 

4-775 

12.272 

59°oo 

35260 

64640 

77570 

30100 

59-7 

10.929 

6.179 

4-750 

12.272 

60420 

37930 

67080 

87260 

36220 

62.7 

10.735 

6.072 

4-663 

12.272 

60420 

38720 

68460 

89150 

33870 

64.0 

11.205 

6.524 

4.681 

12.272 

55520 

36530 

62740 

87440 

36610 

65-8 

11.108 

6.477 

4-631 

12.272 

60430 

38740 

66440 

92920 

35060 

64.1 

... 

9-46S 

5-685 

3.780 

9.818 

59000 

3756o 

62360 

9378o 

36110 

63.6 

9.277 

5-572 

3-705 

9.818 

59000 

39000 

64930 

97650 

36850 

66.1 

•  " 

9-934 

6.119 

3-815 

9.818 

59000 

37600 

61040 

97900 

38040 

63-7 

9-348 

S-758 

3-590 

9.818 

59000 

36000 

58440 

93740 

34280 

61.0 

10.032 

6.322 

3-?™ 

9.818 

59000 

40040 

63540 

108270 

40910 

67-7 

10.187 

6.417 

3-770 

9.818 

5900° 

39720 

63050 

107320 

41210 

67-3 

fracture  occurred  in  the  same  joint   tt   is  indicated    by  employing    larger   type   in   more 
column. 


6i6 


APPLIED  MECHANICS. 


TABLE 
TABLE  OF  EFFICIENCIES  OF 


STEEL  PLATE. 


Plate. 

No.  of 
Test. 

Pitch  of  Rivets. 

it" 

if 

n" 

2" 

**" 

per  cent. 

per  cent. 

per  cent. 

per  cent. 

per  cent. 

1308 

67-5 

68.9 

72.8 

> 

I3I3 

68.3 

69.9 

72.1 

1314 

65-9 

66.4 

6B.'3 

68^7 

i"    .    .    .    .  - 

1323 
1324 

• 

64.8 

67.1 
62.6 

68.6 
64.4 

70.3 
67-5 

^  337 

.... 

.... 

63-9 

65-9 

68.2 

1338 



64.8 

61.7 

1355 

.... 

.... 

.... 

58.7 

60.6 

• 

356 

74-6 

73.3 

72.7 

72.0 

360 

68.2 

69.0 

70.0 

70.9 

t"   •     -     •    -  - 

18 

68.3 

70.1 
64-5 

70-3 
61.7 

68.0 

64.8 

1379 

.... 

63.0 

65-2 

67.4 

1380 



.... 



58.7 

59-6 

• 

1395 



57-3 

59-5 

• 

300 

300 

1396 

.... 

66.1 

79-6 

64.7 



360 

1401 

68.1 

68.6 

68.9 

1402 

.... 

.... 

64-5 

66.3 

400 
76.0 

250 

300 

i"    ....  - 

1411 

.... 

.... 

66.9 

76.4 

66.3 

I4T2 

.... 

.... 



250 
62.8 

59-2 

300 

I42S 

.... 

.... 

60.  1 

68.1 

300 

1426 

.... 

.... 

59-3 

. 

1443 

.... 

5*-S 

1444 

.... 

60.  1 

66.0 

300 

76.2 

i"    .    .    .    .  • 

1451 
1452 

.... 

.... 

62.7 

63.8 
58-3 

66.9 

60.0 

1463 

.... 

... 

58.3 

61.8 

1464 

.... 

56.5 

1481 

.... 

56-5 

1482 

.... 

.... 

54-2 

58.3 

1489 

.... 

54-2 

58.8 

1490 
I5°3 

:::: 

:::: 



52.6 
54-2 

I5°4 

— 

.... 

1523 

•  ... 

.... 

.... 

NOTES.— Figures  in  heavy  face  type  denote  that 
Super  numbers  state  the  temperature  of 


TABULATION  OF  RIVETED  JOINTS. 


6l7 


NO.  i. 

SINGLE-RIVETED  BUTT-JOINTS. 


STEEL  PLATE. 


Pitch  of  Rivets. 

Diam- 
eter of 
Rivet- 
holes. 

2*" 

2f" 

2*" 

2f 

2j" 

2£" 

3" 

3i" 

3i" 

3i" 

per  ct. 

per  ct. 

per  ct. 

per  ct. 

per  ct. 

per  ct. 

per  ct. 

per  ct. 

per  ct. 

perct. 

in. 

$ 
ft 

i 

i 
i 

$ 
I 

i 

i 
i 

1 
1 

i 

i 
i 
i 
i* 
i* 

i 

f 

i 
3 

7.27 
74.0 
70.1 
70.0 
61.5 
63-4 

70.7 
72-3 
63-7 
68.4 

76.8 

67.1 

s; 

69.8 
68.3 
65-4 
68.4 

77.1 
75-0 

73-5 
74.1 

lil 

.... 

.... 

70.7 

68.2 
65-4 
64.1 

72.4 
70.4 
65-3 
68.3 

7S-5 

I1 

69.5 

6&\ 

68.8 
69.8 

73-2 
75.0 

66.4 

600 
77.6 
200 
60.3 

61.5 
360 
67.2 

56.5 

6480°6 

600 

64-3 
63.6 
62.8 

57-8 
58.0 

59.3 

62.3 
58.0 
57-2 
49-3 
50.8 

7o5.°5 
66.7 

400 

72.8 

64-5 
700 
62.7 

58.0 

500 

77-3 
66.9 
62.3 

600 
76.7 

67.2 

400 
76.9 

62.0 
6l.2 

600 
8l.S 

67-6 

eoo 
70.5 

64-5 

600 
80.3 

67.0 

200 
67.5 

70.0 

67.2 
68.2 

.... 

65.8 
65.5 

59-8 
60.7 

64.6 
57.6 

59-1 
60.6 
55  6 
55-7 

66.6 
66.1 

62.5 
62.1 

65.0 
64.5 

66.1 
64.7 

66.5 

66.5 

67.1 

66.6 

68.7 
68.6 

67.9 
69.9 

63  '.6 
62.4 

57-5 
57-4 

67.0 

61.7 
59-7 
61.1 

65-7 
62.2 

=57-0 
59  7 

64-3 
68.8 
62.7 
64.0 

65.8 
64.1 

63.6 

66.1 

66!.J 

67.7 
67.3 

ioint  did  not  fracture  along  line  ot  riveting, 
joints  tested  at  temperatures  above  atmospheric. 


6i8 


APPLIED  MECHANICS. 


TABLE   NO.   2. 

TABLE  OF  DIFFERENCES  BETWEEN  THE  EFFICIENCIES  AND  RATIOS  OF  NET  TO 
GROSS  AREAS. — SINGLE-RIVETED  BUTT-JOINTS,  STEEL  PLATE. 


Plate. 

No.  of 
Test. 

Width  of  Plate  between  Rivet  Holes. 

Diameter 
of  Rivet 
Holes. 

i" 

•i* 

ii" 

if" 

ii" 

i*" 

ii" 

IT 

2" 

in. 
t  • 

t 

* 

I 

f  • 
I 

1  - 

in. 

1308 
1313 
i3M 
I323 
1324 
»337 
1338 
•1355 

1356 
1359 
1360 

1379 
1380 
I39S 

1396 
1401 
1402 
1411 
1412 
1425 
1426 
1443 

1444 

I4SI 

1452 
1463 
1464 
1481 

1482 
1489 
1490 
1503 
1504 
1521 

perct. 

6.0 
6.8 
8.8 

7-7 
8.0 

9-3 
14.8 
18.8 

i3-i 

II.  2 

perct 

4.6 
5-6 

perct. 

6.1 

5-4 

per  ct. 

perct 

perct 

per  ct. 

perct 

perct. 

in. 

i 
* 

I 

i 
i 

* 

j 

i 
i 

i 
i 

i 
i 

i 
ii 
ii 

1 

i 

i 
i 
ii 

ii 

i 
i 
ii 

:i 

ii 

7.0 
8.1 
9.6 
8.8 
7.6 

9.1 
7.7 

6.1 

8.6 
9-4 
5-9 
7.8 

5-6 
11.7 
8.9 
5-8 
10.5 

8.8 
7.6 
9.2 

12.  1 

8-5 

ii.  8 

2.1 

1:1 

1.6 

13-4 

"•3 

8.  '2 
8.8 

'i:? 

II.  2 
II.  2 

U 

7-3 

300 
9-0 

II.  I 

II.  I 

360 
13-5 
360 
12.8 

10.  I 

300 
12.2 

4.6 
6.4 

9-4 

1:1 

9-5 
9-4 

4.1 
4.1 
5-5 

1:1 

6.4 

10.  1 

5-4 
8.9 
6.7 
6.6 

300 

19.6 
8.6 

10.  0 
300 
20.1 

6.2 
300 
15-2 
360 
I7.2 

6-5 

9.7 

7-5 

K 

1:1 

5-3 
5  8 
7-9 

7-2 
8.2 

8-3 

':i 

1:1 

9-9 
5-9 

2.  1 

350 
6.4 

400 
17.2 

7-4 

300 

4-7 

5-9 
700 
10.2 

5-3 

300 

I7-4 

8.1 
8.0 
7.2 

1:1 

8.7 

6.6 
6.4 

7-9 
7-4 
7-3 

3.8 

9-6 
7-i 
7-4 
10.4 

.::: 

9-4 

1:1 

9-5 

10.5 
10.6 

1:1 

*: 

7-9 

9.8 

5-3 

600 

16.4 

400 
14.9 

6.5 
6-9 

600 
21.7 

400 

7-5 

600 

3.2 

8.1 
7.6 
7-5 

7-* 

6-7 
-0.4 
8.6 

1:1 

8.6 

l|6.°8 
3.5 

500 
I7.2 

6.9 

4-9 
4.1 

5-2 
400 
14.9 

600 
II.4 

5-4 

600 
I7.8 

4-0 
600 
19.4 

6.1 

300 

5-o 
7-5 

3.2 
4.2 

6.6 
6.0 

9.0 
7-6 

3.1 
2.7 

7-5 
7-4 

6.2 

5-8 

6.'2 

6.1 

3.9 
5.9 

-.02 

4-7 
2.4 
5-i 

6.6 
3.1 

6.2 

7-4 

3-4 
7-9 
7.6 
5.8 

3.5 

6.0 

2.1 

-0.6 

2i" 

per  ct. 
4.7 
4.3 

NOTES.— Figures  in  heavy-faced  type  denote  that  joint  did  not  fracture  along  line  of  riveting, 
Super  numbers  state  the  temperature    of   joints    tested    at   temperatures    above 
atmospheric. 


TABULATION  OF  RIVETED  JOINTS. 


619 


TABLE   NO.  3. 

EXCESS  IN  STRENGTH  OF  NET  SECTION  IN  JOINT  OVER  STRENGTH  OF  TENSILE 
TEST-STRIP. — SINGLE-RIVETED  BUTT-JOINTS,  STEEL  PLATE. 


Plate, 
in. 
i 

t  ' 

\  • 

I 
1  • 

*  - 
A 

No.  of 
Test. 

Width  of  Plate  between  Rivet  Holes. 

Diameter 
of  Rivet 
Holes. 

x" 

ii" 

ii" 

If" 

4" 

ii" 

If" 

i|" 

2" 

1308 
1313 
1314 
1323 
1324 
1337 
1338 
1355 
1356 
'359 
1360 
i367 
1368 
J379 
1380 

1395 
1396 
1401 
1402 
1411 
1412 
H25 
1426 
1443 
1444 

»45i 
1452 
1463 
1464 
1481 

1482 
1489 
1490 
1503 
i5°4 
1521 

per  ct. 
9.8 
ii.  i 
iS-4 
i3-4 
17.4 
19.7 
29.6 
17.6 

21.2 

18.1 

19-3 
19-5 

20.9 

per  ct. 
7.2 
8.8 
10.5 
11.7 
14.4 
17.0 
16.5 
i4-5 
14.1 
11.9 
15-0 
16.8 
9-7 

per  ct. 
9.1 

8.1 

9-2 

9-7 
14.7 

15-9 
10.6 
14.2 

per  ct. 

perct. 

perct 

perct. 

per  ct. 

perct. 

in. 

f 

i 

1 

i 
i 

* 

I 

i 

i 
i 

i 
$ 

i 
H 
ii 

$ 

i 

i 
i 

'4 

i 
i 
ii 

1 

ij 
ii 

ii- 
i± 

6.2 

8.5 
14.7 
14.6 

10.  0 

18.0 

9-i 

II.  0 

11.9 
14.4 
20.  i 
14.1 

16.1 

IS 

10.7 

4-7 
2.5 

21.0 

17.7 

12.7 
i3-5 

10.6 
10.1 

12.0 
12.5 
10.  1 

9.6 
5.8 
15-6 

14.9 

16.1 

.... 

... 

15.5 
14.6 

200 
15-6 

15-8 

20.9 
350 
25.2 
250 

25-5 

20.3 
800 
26.0 

9.8 
iS-i 

ll:l 

16.6 

2O.  2 
20.0 

8-3 

8.2 

ii.  6 
15-2 
10.8 
14.4 

12.6 

12.5 

300 
32-8 

14.4 

17.8 

800 

35-8 
ii.  8 
238°°8 

350 

34-4 
13-2 

18.8 

13-5 
17.2 
16.8 
15-8 
18.4 

10.  I 

14.1 

16.0 
H-S 
17.4 
17.6 

17.8 
10.6 

3-4 

350 
10.3 
400 
28.9 

12.8 
200 

5-5 
10.8 

700 

12.3 

10.  1 
800 

20.  6 

13.7 

14.6 
13.2 
13-6 
i5-3 
6.8 

12.  1 
I2.3 
15-2 
I5.0 
I4.6 

12.8 

17.9 

16.3^ 
iS-9 

8.9 
10.3 

8.2 

9.8 

12.2 
I5-I 

'.:;: 

8-7 

500 
27.0 

400 

25.8 

"•3 

12.  1 
500 

39-7 

400 
12.2 

600 
S-2 
14.0 
13-2 

13-7 
I3.0 

ii-5 
-0.6 

15.8 
13-6 
13.2 
16.5 

lift 
5.6 

500 
28.7 

"•3 
8-7 
7-3 

\ 

8.6 

400 
24.0 

eoo 
19.4 

9-i 

500 
28.1 

6.4 
500 
32.0 

10.  1 

300 

8.1 

12.2 

5.3 
6.5 

11.1 
10.1 

i5-7 
i3-5 

6.2 
4.3 

12.8 

12.7 

10.2 

9-7 

i'6.'i 

9.8 

6."i 
9.2 

9.S 

8-5 
4-7 
9.6 

II.  2 

5.3 

if.o 

13-3 

5-5 
ii-S 
13.0 
10.0 

5.'  7 

10.  1 

4:» 

«L 

per  ct. 
7.7 
6.9 

verage  of 
all  joints. 

16.2 

14.4 

I2.3 

12.9 

11.9 

II.  0 

9-7 

y 
it.  8 

7.0 

NOTES. — Figures  in  heavy-faced  type  denote  that  joint  did  not  fracture  along  line  of  riveting. 
Super  numbers  state  the  temperature  of    joints  tested    at    temperatures    above 
atmospheric. 


62O 


APPLIED   MECHANICS. 


TABLE    NO.    4. 

COMPRESSION    ON   BEARING   SURFACE    OF    RIVETS    AND    EXCESS  IN  TENSILE 
STRENGTH  OF  NET  SECTION  OF  PLATE. — SINGLE-RIVETED  BUTT- 
JOINTS,  STEEL  PLATE. 


Com- 

Com- 

pression 

pression 

Plate. 

No. 
of 
Test. 

Rivet- 
loles. 

Net 
Width 
of 

Plate. 

on 
Bearing 
Surface 
per 

Excess 
Tensile 
Strength. 

Plate. 

No. 
of 
Test. 

Rivet- 
holes. 

Net 

Width 
of 
Plate. 

on 
Bearing 
Surface 
per 

Excess 
Tensile 
Strength. 

Square 

Square 

Inch. 

| 

Inch. 

in. 

in. 

in. 

Ibs. 

per  ct. 

in. 

in. 

in. 

Ibs. 

per  ct. 

1309 

f 

f  109640 

IS.  I 

1404 

i 

1 

86670 

17  8 

1308 

| 

108370 

9.8 

H 

1414 

i 

75430 

ii.  8 

* 

94890 

15-4 

1 

1429 

it 

66840 

13.2 

i- 

1315 

i 

93330 

13-4 

1325 

I 

84510 

19.7 

1324 

1 

92870 

r 

1447 

£ 

80280 

13-5 

1339 

i 

72580 

\]'.l 

1454 

I 

72510 

17.2 

i] 

1455 

I 

72250 

16.8 

1356 

i 

105170 

21.2 

1466 

It 

}-  it 

1  66240 

15-8 

1357 
1361 

I 

102490 
86430 

18.1 

19-5 

I 

1467 

:  66250 

18.4 

f- 

1360 
1368 
1369 
1380 
1381 

i 

I 

i 
i 

86330 
75000 
73150 
69010 
66880 

19.3 

20.9 
18.2 

'5-5 
14.6 

r 

i 

1484 
1492 
H93 
1507 
1506 

I 
It 

73200 
68640 
67520 
63000 
L  63890 

10.  I 

16.0 

14  i 
17.6 

17.4 

r 

!397 

I 

»•  i 

-  90960 

iS-8 

J 

1402 

£ 

7913° 

20.9 

r 

1312 

1 

"I 

r  134750 

9-1 

1 

1413 

i 

68780 

20.3 

1313 

§ 

133720 

8.1 

I 

1427 

it 

57330 

9.8 

1319 

i 

11475° 

9-7 

a.. 

1318 

f 

112610 

9-2 

r 

1445 

i 

74000 

17.7 

1 

1329 

I 

103860 

15-9 

1444 

* 

7233° 

jc  m  i 

1328 

i 

102810 

14.7 

J 

1452 

64150 

16.6 

1343 

i 

87750 

14.2 

§1 

i 

64110 

16.6 

I 

1342 

X 

85130 

10.6 

1465 

it 

58780 

20.  o 

I 

1464 

it 

58690 

20.2 

r 

1365 

I 

1  12000 

12.5 

f. 

1491 
1482 
1483 
1490 

it 

i 
i 
it 

65340 
64090 
64070 
58750 

15.2 

8-3 
8.2 

ii.  6 

i 

1364 
1373 

1384 

f 

! 
i 
i 

III5IO 
94580 
94210 
80570 
79810 

12.0 

13-5 
IO.  I 

10.6 
17.8 

1505 

54010 

14.4 

i£ 

I  52460 

10.8 

•  it 

.   ' 

r 

1400 

{• 

98830 

3-4 

r 

1311 

i 

f  119810 

8.8 

J 

1407 

1 

95310 

12.8 

1310 

119180 

7-2 

1 

1417 

i 

83220 

10.8 

1317 

f 

103750 

ii.  7 

I 

1431 

it 

71160 

10.  1 

J 

1316 

f 

102630 

10.5 

*1 
I 

1327 
1326 
1340 
i34i 

i 

i 
i 

•  it 

9440*0 
92210 
82150 
-  80820 

17.0 
14.4 
16.5 

§! 

1456 
H57 
1469 
1468 

i 

it 
it 

82IIO 
81040 
73540 
72330 

14.6 

13.2 

'5-3 
13.6 

1363 

j 

95  1  3° 

16.8 

1362 

I 

93620 

15.0 

f 

.1487 

i 

82870 

12.  1 

89000 

i5-9 

M95 

it 

75650 

15-2 

137° 

Q 

84280 

9-7 

i  i 

1494 

it 

73780 

12.3 

1382 

i 

75640 

12.6 

1508 

'4 

69650 

15.0 

1383 

1 

J 

I  738oo 

12.5 

1 

iS°9 

ii 

J 

L  69420 

!4.6 

TABULATION  OF  RIVETED  JOINTS. 


621 


COMPRESSION  ON  BEARING  SURFACE  OF  RIVETS,  ETC  — Continued. 


Com- 

Com- 

pression 

pression 

Plate. 

No. 
of 
Test. 

iivet- 
loles. 

Net 
Width 
of 
Plate. 

on 
Bearing 
Surface 
per 

Excess 
Tensile 
Strength. 

Plate. 

No. 
of 
Test. 

Rivet- 
holes. 

Net 
iVidth 
of 

Plate. 

on 
Bearing 
Surface 
per 

Excess 
Tensile 
Strength. 

Square 

Square 

Inch. 

Inch. 

in. 

in. 

in. 

Ibs. 

per  ct. 

in. 

in. 

in. 

Ibs. 

per  ct. 

1321 

| 

f  124880 

8-5 

f 

1334 

f 

12755° 

16.1 

1320 

f- 

122140 

6.2 

|j 

1335 

1 

123030 

3-2 

133° 

Z 

110830 

14.7 

1 

1349 

I 

112380 

10.7 

t' 

1331 

j 

106620 

14..  6 

I 

1348 

I 

101620 

5-6 

1345 

i 

96070 

18.0 

1344 

i 

92870 

IO.O 

r 

1379 

1 

116280 

16.5 

«1 

1378 

115680 

16.1 

{ 

1366 

f 

116340 

9.6 

1 

i 

103690 

10.3 

1375 

i 

102360 

n-7 

I 

1390 

i 

102340 

8.9 

H 

1374 

I 

97620 

15-6 

)•  if 

1386 

87910 

12.8 

ij 

1422 

i 

104490 

8.6 

I 

1387 

i 

87350 

17.9 

M 

1437 

ii 

9473° 

9.1 

I 

1408 

$ 

•  if 

102500 

8.7 

i 

1475 

ii 

93120 

12.7 

M 

1419 

92150 

ii  -3 

1 

1474 

xi 

93110 

12.8 

( 

H32 

ii 

81430 

12.  I 

1500 

ii 

9475° 

II.  2 

J 

1458 
M59 

5 

89260 
89170 

14.0 
13.2 

1515 

i 

J 

89150 
87260 

13-3 
II.  0 

f  1 

1470 

xi 

77740 

13-7 

I 

1471 

i| 

77270 

I3.0 

\ 

1337 

I 

c  125990 

2.7 

iJ 

1336 

i 

124020 

4-7 

'V 

1488 

I 

90600 

ii  5 

1 

i 

118300 

31.0 

1496 

ii 

85480 

15-8 

I 

1351 

i 

114980 

17.7 

ij, 

1497 

83980 

13-6 

t 

1510 

i 

77710 
I  76680 

16.5 

13.2 

»i 

1393 
1392 

! 

i  i  1070 
109540 

9.8 

8.2 

1323 

j 

1 

131460 

ii  .0 

,  $ 

1425 

i 

•  if 

-  111490 

6.4 

1322 

j. 

129150 

9.0 

i 

1439 

ii 

99450 

10.  1 

, 

*" 

1333 

I 

116070 

14.4 

1332 

i 

113500 

11.9 

4  -I 

1476 

Ji 

99410 

10.2 

1346 

i 

108960 

20.  i 

1 

1477 

xi 

97430 

9-7 

1347 

i 

103540 

14.1 

j 

1502 

ii 

97050 

5-5 

f 

1377 

f 

108260 

11.4 

H 

1503 

Ji 

90600     1  1  .  5 

f  J 

i376 

f 

105470 

14.9 

! 

1516 

ii 

J 

I  87440 

13.0 

1 

1388 

i 

937io 

16.3 

1 

1389 

i 

.  T! 

.  93390 

iS-9 

,  j 

1353 

i 

f  126160 

13-5 

IT 

I 

1352 

i 

125220 

12.7 

i 

1421 

i 

96850 

n-3 

H 

1434 

ii 

85540 

8-7 

,  j 

1395 

i 

115660 

15.1 

t 

1435 

84420 

7-3 

f1 

1394 

i 

}•  ii 

j  113240 

12.2 

j 

1473 

ii 

86400 

13-5 

% 

1441 

ii 

108220 

12.2 

1 

1472 

ii 

84770 

i 

l| 

1  102440 

9.8 

t 

1519 

ii 

J 

I  97650 

10.  1 

i 

1499 
1513 

i* 

84810 
77570 

8-5 
9.6 

I 

1354 

i 

•  2 

j  133730 

10.6 

I 

1512 

* 

I  74320 

4-7 

1 

1520 

l} 

]  97900 

3-5 

622  APPLIED  MECHANICS. 

RIVETED   JOINTS. 

TESTS    MADE    FOR    THE    BUREAU    OF    STEAM-ENGINEERING,  U.  S.  NAVY 

DEPARTMENT. 

The  specimens  were  prepared  at  the  Brooklyn  Navy-yard. 

The  joints  were  made  of  open-hearth  steel  plates,  which  had 
drilled  rivet-holes  and  sheared  edges.  Both  steel  and  wrought- 
iron  rivets  were  employed,  as  noted  in  the  details.  They  were 
machine-driven. 

The  strips  tested  to  show  the  quality  of  the  metal  employed 
in  the  joints  were  furnished  by  the  steel-makers  as  being  the 
same  grade  of  metal,  although  not  from  the  same  sheets  as 
the  joints. 

The  tests  were  conducted  in  the  same  manner  as  earlier 
tests  of  this  class  recorded  in  previous  reports. 

The  gauged  length  was  established  symmetrically  with  the 
centre  line  of  the  joint,  and  observations  were  made  on  this 
section  at  intervals  during  the  progress  of  the  test. 

It  will  be  seen  that  the  thickness  of  the  plates  varied  con- 
siderably at  different  edges ;  these  measurements  are  in  their 
respective  places  on  the  sketches  which  accompany  the  details. 

After  testing  from  a  number  of  the  joints,  the  rivet-heads 
were  planed  off,  the  rivets  driven  out,  and  butt-straps  removed, 
thus  giving  access  to  the  rivet-holes  in  the  plate,  which  were 
then  measured  and  their  elongation  recorded  on  the  sketch. 

From  these  figures  the  transmission  of  stresses  to  the  differ- 
ent rows  of  rivet-holes  may  be  approximately  judged  of. 

The  metal  drew  down  in  thickness  in  zigzag  lines  from  rivet- 
hole  to  rivet-hole  of  adjacent  rows,  notwithstanding  the  extent 
of  metal  thus  traversed  was  greater  than  directly  across  the 
plate  through  one  line  of  rivet-holes. 

The  average  tensile  strength  of  the  three  specimens  of  each 
thickness  of  plate  was  used  in  computing  the  efficiencies  of  the 
joints. 


TABULATION  OF  RIVETED  JOINTS. 


623 


8   £ 

2§ 

s  e 

a  fc 


-       N          ^ 


ky, 
am 


g 
. 


ky, 
am 


y          N 

I    "' 


;i 

^ 


W  .2 


5    ij 


•2< 


•ON  jsax 


624 


APPLIED   MECHANICS.. 


TABULATION   OF   RIVETED   JOINT.S    FOR    UNITED   STATES 


Nominal 

No.  of 
Test. 

Marks. 

Style  of  Joint. 

Thick- 
ness of 

Size  and  Kind  of 
Rivets  and  Holes. 

Plate. 

910 

A, 

3"pitch. 

inch. 

911 

A2 

M 

D°o<ifo0o0Q0< 

¥^oWo< 

LJJL?^_ 

-»7.35  x  .50 

I 

f-inch  steel  rivets,  §f- 
inch  drilled  holes. 

912 

AS 

t 

20.00 

* 

i 

Switch. 

9I3 

i 

>:£)   O   O   O  O 

1 

f 

•D 

ij 

G§QOQ°O°O(~ 

^     H 

f-inch  steel  rivets,  §|- 

914 

Dg 

c 

orf)  ;OL  o  o 

li 

inch  drilled  holes. 

QTC 

Bo 

v 

O   O   Q   O   O 

T 

yxo 

JJ3 

20.00 

•> 

916 

D 

3^"  pitch. 

•^ 

j 

f 

Q  O  O  O'  O 

t 

WS 

i 

917 

D2 

2^50  o  o 

Hi 

j 

.  i-inch  steel  rivets,  i^g-  J 

018 

o°o®o0o°q 

1 

I 

i 

inch  drilled  holes. 

yA° 

3 

«—              J7.00              "* 

8 

919 

Ej 

4%6"pitch. 

| 

r 

C§r  O    O'    O 

^TO  vQ    O 

a 

920 

E8 

g)  ^  _Q_.jQ 

i 

i 

i-inch  stee/  rivets,  i^-  J 
inch  drilled  holes,      j 

r*\     r\     f*\ 

CO 

r^ 

O    O    O    O 

1 

921 

3 

t_             16.50            -» 

8 

922 

G, 

O    2%'pitch.  O 

i 

3  plates.               f 
i-inch    iron    rivets, 
i^-inch  drilled  holes.   | 

923 

G2 

I 

c£?o^°o| 

| 

<* 

5 

i 

2  plates. 

i^-mch  iron  rivets,  i£-  ' 

5? 

RJC^^OT^' 

inch  drilled  holes. 

o      5*»J5 

924 

G3 

O         '€)   i 

f 

i 

J 

IT             15,75 

TABULATION   OF  RIVETED  JOINTS. 


625 


NAVY   DEPARTMENT,  BUREAU    OF   STEAM-ENGINEERING. 


Sectional  Area 
of  Plate. 

Bearing 
Surface 
of 
Rivets. 

Shear- 
ing 
Area  of 
Rivets. 

Tensile 
Strength 
of  Plate, 
pounds 
per 
Square 
Inch. 

Maximum  Stress  on  Joint  per  Sq.  In. 

Effici- 
ency of 
Joint. 

Tension 
on  Gross 
Section  of 
Plate. 

Tension 
on  Net 
Section  ol 
Plate. 

Compres 
sion  on 
Bearing 
Surface 
of  Rivets 

Shear- 
ing on 
Rivets. 

Gross. 

Net. 

sq.  in. 

sq.  in. 

sq.  in. 

sq.  in. 

Ibs. 

Ibs. 

Ibs. 

Ibs. 

per  ct. 

13.22 

IO.  12 

6.72 

12.46 

53710 

42860 

55990 

84320 

45470 

79-8 

13.08 

10.01 

6.64 

12.46 

537" 

40960 

53530 

80690 

43000 

76.2 

12.41 

9-5° 

6.30 

12.46 

537io 

42720 

55800 

84140 

42540 

79-5 

12.81 

10.31 

8.01 

iS'34 

53710 

43460 

54040 

69560 

36320 

80.9 

12.69 

IO.2I 

7-93 

15-34 

537io 

40390 

50200 

64630 

334io 

75-2 

12.691 

10.21 

7-94 

15-34 

537io 

44290 

55050 

70790 

36640 

82.5 

14.671 

lO.og 

8.25 

15.96 

51190 

35180 

51150 

62560 

32340 

68.7 

14.646 

IO.O7 

8.23 

15.96 

,     5"9o 

36190 

52640 

64410 

33210 

70.7 

J4-705 

IO.II 

8.27 

15.96 

51190 

3578o 

52050 

63630 

32970 

69.9 

14.322 

10.63 

10.15 

19-51 

,5H9o 

379io 

51080 

535oo 

27830 

74-1 

14.438 

10.72 

10.23 

19-51 

51190 

38400 

51720 

54i9o 

28420 

75-0 

14-256 

10.58 

10.10 

I9-5I 

51190 

37950 

51130 

5356o 

27730 

74-1 

13.781 

10.93 

16.30 

28.00 

51190 

40810 

51460 

34500 

20090 

79-7 

13-852 

10.985 

16.38 

28.00 

51190 

41740 

52640 

35300 

20650 

81.5 

I3.74I 

10.90 

16.26 

28.00 

51190 

40120 

50580 

339io 

19690 

78.4 

(The  figures  in  heavy-faced  type  indicate  the  manner  of  failure.] 

626 


APPLIED  MECHANICS. 


TABULATION  OF  RIVETED  JOINTS  FOR  UNITED  STATES  NAVY 


No.  of 
Test. 

Marks. 

Style  of  Jotot. 

Nominal 
Thick- 
ness of 
Plate. 

Size  and  Kind  of 
Rivets  and  Holes. 

inch. 

925 

H, 

I 

I 

f 

3  plates, 
i-inch  iron  rivets,  i^g- 
inch  drilled  holes. 

Qwi&r^^^P 

^j  -.    w  ,- 

fl    ^J        f2%  pitch.,       ^ 

926 

* 

9 

1 

wo°o^m 

s 

* 

2  plates. 
i£-inch  iron  rivets,  ij-  T 
inch  drilled  holes. 

°rfsm% 

6_  —  ~—  «—  —  -g> 
M?l 

'-V 

927 

H3 

1 

028 

I. 

A 

q^**^» 

1 

ro~A"6~b~c 

929 

i» 

*H 

8         y 

A 

If-inch  iron  rivets,  i-  i 
'     inch  drilled  holes. 

t 

)  o  6  Q  o  c 

a 

93° 

.3 

0        jfk    0 

A 

*—         14.38          *^» 

8^16  pitch. 

93  * 

* 

o  o  o  o  o 

f 

932 

K3 

i 

f 

!-inch  iron  rivets,  f§- 
\-    inch  drilled  holes,     j 

933 

20.00 

TABULATION   OF  RIVETED  JOINTS. 


627 


DEPARTMENT,    BUREAU    OF    STEAM-ENGINEERING—  Continued. 


Sectional  Area 
of  Plate. 

Bearing 
Surface 
of 
Rivets. 

Shear- 
ing 
Area  of 
Rivets. 

Tensile 
Strength 
of  Plate, 
per 
Square 
Inch. 

Maximum  Stress  on  Joint  per  Sq.  In. 

Effici- 
ency of 
Joint. 

Tension 
on  Gross 
Section 
of  Plate. 

Tension 
on  Net 
Section 
of  Plate. 

Compres- 
sion on 
Bearing 
Surface 
of  Rivets. 

Shear- 
ing on 
Rivets. 

Gross. 

Net. 

sq.  in. 

sq.  in. 

sq.  in. 

sq.  in. 

Ibs. 

Ibs. 

Ibs. 

Ibs. 

Ibs. 

per  ct. 

13-252 

10-537 

14-045 

30.41 

537io 

42820 

53860 

40410 

18660 

79-7 

12.776 

10.154 

13-548 

30-41 

537" 

453oo 

57000 

42720 

19030 

84.3 

12.  8l 

10.186 

13-576 

30.41 

.  537io 

46070 

57940 

43470 

19410 

85.8 

8.287 

6.820 

8.057 

18.06 



42250 

51330 

43450 

19380 

.... 

8.  211 

6-755 

7-994 

18.06 



40800 

49590 

41910 

18550 

.... 

8.28 

6.81 

8.05 

18.06 



40720 

49520 

41890 

18670 

.... 

12.36 

9.94 

7-73 

15-34 

537io 

43600 

54220 

69720 

35130 

81.2 

12.93 

10.40 

8.08 

15-34 

537io 

43260 

53780 

69220 

36460 

80.5 

12.98 

10.44 

8.  ii 

15.34 

537io 

43000 

5346o 

68820 

36390 

80.  i 

628 


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TABULATION  OF  RIVETED  JOINTS. 


629 


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RIVETED  JOINTS. 


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632  APPLIED   MECHANICS. 

GENERAL   REMARKS   ON   RIVETED   JOINTS. 

In  designing  a  single-riveted  joint,  one  of  the  first  matters  to  be  de- 
cided is  the  size  of  rivet.  In  this  regard  we  have  first  certain  rules, 
which  may  be  said  to  be  the  common  custom.  Now,  if  the  strength  per 
square  inch  of  the  plate  between  the  holes  remained  constant  for  all 
pitches  and  sizes  of  rivets,  it  is  plain  that,  inasmuch  as  the  larger  the 
rivets  the  larger  the  pitch,  and  as  the  rivet-area  increases  with  the 
square  of  the  diameter,  whereas  the  plate  increases  with  the  first  power 
of  the  length,  it  would  follow  that  the  larger  the  rivets  the  higher  the 
efficiency  which  we  might  expect.  The  fact  that  the  excess  tenacity  of 
the  plate  between  the  holes  is  greater  with  small  than  with  larger  dis- 
tances between  the  holes  comes  in  to  modify  the  above  action  to  some 
extent;  but  what  little  evidence  there  is  from  experiment  would  seem 
to  point  still  to  there  being  an  advantage  in  using  larger  rivets  than  is 
usually  done.  Thus  Kennedy's  tests  indicate  this,  though  they  were 
made  on  narrow  strips ;  and  also  a  few  of  the  Government  tests  were 
made  with  the  same  section  of  metal  between  the  holes,  and  different 
sizes  of  rivets;  and  the  advantage  seems  to  be  with  the  larger  rivets. 
The  difficulties  that  are  liable  to  present  themselves  in  regard  to  large 
rivets  and  large  pitches  are:  i°.  They  must  not  be  so  large  as  to  make 
it  impossible  for  the  riveter  to  secure  a  well-filled  hole  and  a  tight  joint. 
2°.  In  boiler-work  the  pitch  must  not  be  so  large  as  to  render  it  uncer- 
tain whether  the  joint  can  be  made  steam-tight  by  calking. 

If  we  could  assume  a  fixed  number  as  the  strength  per  square  inch 
of  the  plate  as  holding  for  all  sizes  of  rivets  and  all  pitches,  we  should 
merely  fix  upon  our  size  of  rivet  and  then  determine  the  pitch  in  such  a 
way  as  to  make  the  joint  equally  strong  to  resist  tearing  or  shearing ;  but 
as  we  must  take  into  consideration  the  diminution  in  the  excess  tenacity 
of  the  plate  as  the  distance  between  holes  increases,  we  shall  have  to 
consult  the  tests  themselves  in  fixing  upon  both  the  size  of  rivet  and  the 
pitch  if  we  wish  the  most  efficient  joint  under  the  circumstances. 

Of  course  in  a  lap-joint  the  bending  of  both  plates  and  rivets  comes 
in  to  affect  the  strength,  and,  to  guide  us  to  know  h6w  much,  we  can 
refer  to  the  earlier  tests. 

Then,  again,  in  the  case  of  boiler-work,  we  ought  to  take  into 
account  the  fact  that  the  plate  is  almost  sure  to  corrode  and  get  thinner 
by  use,  while  the  rivet  is  not  liable  to  do  so. 

In  double-riveting  similar  remarks  apply,  though,  as  we  have  very 
few  reliable  tests  upon  this,  and  nothing  systematic,  no  more  will  be 
said  here. 


CHAIN   CABLE.  633 


§  233.  Chain  Cable.  —  The  most  thorough  set  of  tests  of  the 
strength  of  chain  cable  is  that  made  by  Commander  Beardslee 
for  the  United-States  government,  an  account  of  which  may  be 
found  either  in  the  report  already  referred  to,  or  in  the  abridg- 
ment by  William  Kent. 

In  this  report  are  to  be  found  a  number  of  conclusions, 
some  of  which  are  as  follows  :  — 

i°.  That  cables  made  of  studded  links  (i.e.,  links  with  a 
cast-iron  stud,  to  keep  the  sides  apart)  are  weaker  than  open- 
link  cables. 

2°.  That  the  welding  of  the  links  is  a  source  of  weakness ; 
the  amount  of  loss  of  strength  from  this  cause  being  a  very 
uncertain  quantity,  depending  partly  on  the  suitability  of  the 
iron  for  welding,  and  partly  on  the  skill  of  the  chain-welder. 

3°.  That  an  iron  which  has  a  high  tensile  strength  does  not 
necessarily  make  a  good  iron  for  cables.  Of  the  irons  tested, 
those  that  made  the  strongest  cables  were  irons  with  about 
5 1000  Ibs.  tensile  strength. 

4°.  The  greatest  strength  possible  to  realize  in  a  cable  per 
square  inch  of  the  bar  from  which  it  is  made  being  200  per 
cent  of  that  of  the  bar-iron  from  which  it  was  made,  the  cables 
tested  varied  from  155  to  185  per  cent  of  that  of  the  bar- 
iron. 

5°.  The  Admiralty  rule  for  proving  chain  cables,  by  which 
they  are  subjected  to  a  load  in  excess  of  their  elastic  limit, 
is  objected  to,  as  liable  to  injure  the  cable :  and  the  report 
suggests,  in  its  place,  a  lower  set  of  proving-strengths,  as  given 
in  the  following  table ;  the  Admiralty  proving-strengths  being 
also  given  in  the  table. 

In  these  recommendations,  account  is  taken  of  the  different 
proportion  of  strength  of  different  size  bars  as  they  come  from 
the  rolls,  also  no  proving-stress  is  recommended  greater  than 
50  per  cent  of  the  strength  of  the  weakest  link,  and  45.5  per 
cent  cf  the  strongest ;  whereas  in  the  Admiralty  tests,  66.2 


634 


APPLIED  MECHANICS. 


per  cent  of  the  strength  of  the  weakest,  and  60.3  per  cent  of 
the  strongest,  is  sometimes  used. 

For  the  details  of  this  investigation,  see  the  report,  Execu- 
tive Document  No.  98,  45th  Congress,  second  session,  or  the 
abridgment  already  referred  to. 


Diameter  of 

Recommended 

Admiralty 

Diameter  of 

Recommended 

Admiralty 

in  inches. 

Proving-Strains. 

Proving-Strains. 

Iron, 
in  inches. 

Proving-Strains. 

Proving-Strains. 

2 

121737 

161280 

tfr 

66138 

83317 

lit 

114806 

I5I357 

if 

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76230 

* 

108058 

I4I750 

!-rV 

55903 

69457 

lit 

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132457 

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63000 

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123480 

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51030 

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106470 

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455J7 

i* 

77159 

98437 

i 

33840 

40320 

ii 

71550 

90720 

§  234.  Iron  and  Steel  Wire.  —  It  has  long  been  known  that 
the  process  of  cold-drawing,  by  which  wrought-iron  and  steel  are 
made  into  wire,  greatly  increases  its  strength  per  square  inch  ; 
and  usually  the  smaller  the  wire,  the  greater  the  strength  per 
square  inch.  Thus  iron  wire  varies  in  strength  from  about  70000 
to  about  90000  Ibs.  per  square  inch  when  unannealed,  its  strength 
being  reduced  to  45000  or  50000  Ibs.  by  being  annealed. 

In  steel  wire  unannealed,  the  strength  runs  up  even  to 
200000  Ibs.  per  square  inch  in  some  cases ;  and  Fairbairn  even 
records  strengths  as  great  as  275000  Ibs.  per  square  inch. 

Accounts  of  experiments  upon  the  strength  and  elasticity 
of  wrought-iron  and  steel  wire  will  be  found  in  — 

W.  E.  Woodbridge  :  Report  on  the  Mechanical  Properties  of  Steel. 
Professor  Thurston  :  Materials  of  Engineering  and  other  papers. 
Pocket-Book  of  the  New-Jersey  Iron  and  Steel  Company. 


IRON  AND   STEEL    WIRE. 


635 


The  strength  of  wire  becomes  of  special  importance  in 
connection  with  wire  rope,  and  also  with  wire-wound  guns. 
Dr.  Woodbridge  has  made  a  great  number  of  tests  of  steel  wire 
for  wire-wound  guns. 

Wire  Rope.  —  Inasmuch  as  wire  rope  is  extensively  used  in 
the  transmission  of  power,  it  may  be  a  matter  of  convenience 
to  have  here  a  table  giving  the  strength  of  the  rope  as  claimed 
by  some  of  the  makers.  There  will  follow,  therefore,  the  table 
of  strengths  of  the  different  sizes  as  given  by  Mr.  John  A, 
Roebling  for  the  rope  manufactured  by  him,  and  also  some  o£ 
his  remarks  in  regard  to  its  use :  — 


Rope  of  133  Wires,  or  19  Wires  to  the 
Strand. 


Rope  of  49  Wires,  or  7  Wires  to  the 
Strand. 


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17280 

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io* 

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ioi 

If 

8540 

4 

4 

6960 

.s 

i 

•8J 

| 

g 

f 

1 

slf 

I 

2n 

Q  P^  y£ 

1 

P 

•5  c 

1  sl 

ill 

b 

ii 

4f 

72000 

lOf 

12 

4i 

60000 

10 

13 

3f 

50000 

9i 

14 

3t 

40000 

8i 

'15 

3 

32000 

7i 

16 

2f 

24600 

61- 

17 

2f 

16600 

Si 

18 

2i 

15200 

5 

19 

Ig- 

11600 

4t 

20 

If 

8180 

4 

21 

if 

5660 

3i 

22 

i| 

4260 

2! 

23 

ii 

3300 

2i 

24 

i 

2760 

2i 

25 

| 

2060 

2 

26 

1 

1620 

if 

27 

f 

1  120 

H 

'636  APPLIED  MECHANICS. 

Notes  by  Mr.  Roebling.  —  "Two  kinds  of  wire  rope  are 
manufactured.  The  most  pliable  variety  contains  nineteen 
wires  in  the  strand,  and  is  generally  used  for  hoisting  and 
running  rope.  The  ropes  with  twelve  wires  and  seven  wires 
in  the  strand  are  stiffer,  and  are  better  adapted  for  standing- 
rope,  guys,  and  rigging.  Ropes  are  made  up  to  three  inches  in 
diameter,  both  of  iron  and  steel,  upon  special  application. 

"  For  safe  working-load,  allow  one-fifth  to  one-seventh  of  the 
ultimate  strength,  according  to  speed,  so  as  to  get  good  wear 
from  the  rope.  When  substituting  wire  rope  for  hemp  rope,  it 
is  good  economy  to  allow  for  the  former  the  same  weight  per 
foot  which  experience  has  approved  for  the  latter. 

"  Wire  rope  is  as  pliable  as  new  hemp  rope  of  the  same 
strength  :  the  former  will  therefore  run  over  the  same  size 
sheaves  and  pulleys  as  the  latter.  But  the  greater  the  diameter 
of  the  sheaves,  pulleys,  or  drums,  the  longer  wire  rope  will 
last.  In  the  construction  of  machinery  for  wire  rope,  it  will  be 
found  good  economy  to  make  the  drums  and  sheaves  as  large 
as  possible. 

"  Experience  has  demonstrated  that  the  wear  increases  with 
the  speed.  It  is  therefore  better  to  increase  the  load  than  the 
speed. 

"  Wire  rope  is  manufactured  either  with  a  wire  or  a  hemp 
•centre.  The  latter  is  more  pliable  than  the  former,  and  will 
wear  better  where  there  is  short  bending. 

"Wire  rope  must  not  be  coiled  or  uncoiled  like  hemp  rope. 
When  mounted  on  a  reel,  the  latter  should  be  mounted  on  a 
spindle  or  flat  turn-table,  to  pay  off  the  rope.  When  forwarded 
in  a  small  coil  without  reel,  roll  it  over  the  ground  like  a  wheel, 
and  run  off  the  rope. in  that  way.  All  untwisting  or  kinking 
must  be  avoided." 

In  the  case  of  wire  rope  it  is  true,  as  in  all  other  cases,  the 
only  way  to  secure  certainty  in  regard  to  the  strength  is  to  test 
it.  The  author  has  tested  wire  rope  which  bore  no  more  than 


OTHER   METALS  AND  ALLOYS.  637 

two-thirds  the  strength  claimed  for  it ;  though  it  is  but  justice 
to  say,  in  this  connection,  that  the  few  samples  of  Roebling 
rope  tested  in  his  laboratory  bore  out  very  fairly  the  results 
given  for  the  sizes  tested  in  the  table. 

A  number  of  tests  of  wire  rope  have  been  made  on  the 
government  testing-machine  at  Watertown  Arsenal  and  else- 
where. 

Wire  rope  is  generally  used  with  some  kind  of  holder,  and 
it  is  generally  the  case  that  the  breakage  of  the  rope  occurs  at 
the  holder. 

Another  matter  worth  noticing  is,  that  it  yields  in  detail, 
and  hence  that  new  rope  is  not  as  strong  as  that  which  has 
been  under  tension  for  some  time,  provided  the  tension  has  not 
been  excessive, 

§235.  Other  Metals  and  Alloys.  —  Copper  is,  next  to 
iron  and  steel,  the  metal  most  used  in  construction,  sometimes 
in  the  pure  state,  especially  in  the  form  of  sheets  or  wire,  but 
more  frequently  alloyed  with  tin  or  zinc ;  those  metals  where 
the  tin  predominates  over  the  zinc  being  called  bronze,  and 
those  where  zinc  predominates  over  tin,  brass.  Copper  in  the 
pure  state  was  used  not  long  ago  for  the  fire-box  plates  of  loco- 
motive and  other  steam-boilers,  as  it  was  believed  to  stand  better 
the  great  strains  due  to  the  changes  of  temperature  that  come 
upon  these  plates,  than  iron  or  steel ;  but  now  steel  or  iron  has 
almost  entirely  superseded  it  for  this  purpose,  except  in  some 
cases  where  the  feed-water  is  very  impure,  and  where  the 
impurities  are  such  as  corrode  iron. 

The  alloys  of  copper,  tin,  and  zinc  which  are  used  most 
where  strength  and  toughness  are  needed,  are  those  where  the 
tin  predominates  over  the  zinc ;  and  the  composition,  mode  of 
manufacture,  and  resisting  properties  of  these  metals,  together 
with  the  effect  of  other  ingredients,  as  phosphorus,  have  been 
very  extensively  investigated  with  reference  to  their  use  as  a 
material  for  making  guns,  instead  of  cast-iron. 


638 


APPLIED   MECHANICS. 


Accounts  of  tests  made  on  these  alloys  will  be  found  as 
follows :  — 

Major  Wade:  Ordnance  Report,  1856. 

T.  J.  Rodman :  Experiments  on  Metals  for  Cannon. 

Executive  Document  No.  23,  46th  Congress,  26.  session. 

No  attempt  will  be  made  to  give  a  complete  account  of  the 
results  of  these  tests ;  but  a  table  will  be  given  for  convenience 
of  use,  showing  rough  average  values  of  the  resisting  powers  of 
some  metals  and  alloys  other  than  iron. 


Specific 
Gravity. 

Tensile 
Strength 
per  Sq.  In. 

Modulus 
of 
Elasticity. 

8.^96 

ISOOO 

9170000 

49000 

14230000 

Bronze  unwrought  : 

84.29  copper  +  15.71  tin  (gun  metal) 

8.561 

36060 

- 

82.81       "      +  17.19  "           " 

8.462 

34048 

- 

81.10      "      +  18.90  " 

8-459 

39648 

- 

78.97       "      +21.03  "  (brasses)  .    .    . 

8.728 

30464 

- 

34.92      "      +65.08  "  (small  bells)     . 

8.056 

3J36 

- 

15.17       "      +84.83  "  (speculummetal) 

7-447 

6944 

- 

Tin  

7.2QI 

t;6oo 

_ 

Zinc     

6.861 

7  «x> 

_ 

8.712 

24138 

_ 

8.878 

T?ooo 

_ 

Copper  wire           ......... 

60000 

I7OOOOOO 

Gold  cast  .     

IQ.2C8 

2OOOO 

10.4.76 

40000 

22.069 

56000 

11.^^2 

1800 

Professor  Thurston  gives,  for  the  ultimate  tensile  strength 
per  square  inch  of  a  compound  of  copper  and  tin  and  zinc, 

T  =  30000  +-  iooo/  +  5002, 


TIMBER.  639 


where  /  —  percentage  of  tin,  and  not  over  15  per  cent;  and 
z  =  percentage  of  zinc,  and  not  over  50  per  cent. 

Some  specimens  of  phosphor  bronze  tested  by  Kirkaldy 
gave  for  tensile  strengths  from  22000  to  50000  for  cast,  and 
98000  to  159000  for  phosphor  bronze  wire  0.06  inch  to  o.  11 
inch  diameter.  If  the  student  is  to  use  alloys,  he  should  ascer- 
tain their  strength,  elastic  limit,  and  modulus  of  elasticity,  and 
he  should  observe  that  the  mode  of  manufacture  has  a  great 
influence  on  the  strength  and  ductility  of  any  alloy. 

§  236.  Timber.  —  However  extensively  iron  and  steel  may 
have  superseded  timber  in  construction,  nevertheless,  there  are 
many  cases  in  which  iron  is  entirely  unsuitable,  and  where 
timber  is  the  only  material  that  will  answer  the  purpose ;  and 
in  many  cases  where  either  can  be  used,  timber  is  much  the 
cheaper.  Hence  it  follows  that  the  use  of  timber  in  construc- 
tion is  even  now,  and  as  it  seems  always  will  be,  a  very  impor- 
tant item. 

Another  advantage  possessed  by  timber  is,  that,  on  yielding, 
it  gives  more  warning  than  iron,  thus  affording  an  opportunity 
to  foresee  and  to  prevent  accident. 

If  we  make  a  section  across  any  of  the  exogenous  trees,  as 
the  oak,  pine,  etc.,  we  shall  find  a  series  of  concentric  layers  ; 
these  layers  being  called  annual  rings,  because  one  is  generally 
deposited  every  year. 

Radiating  from  the  heart  outwards  will  be  found  a  series  of 
radial  layers,  these  being  known  as  the  medullary  rays. 

Of  the  annual  rings,  the  outer  ones  are  softer  and  lighter  in 
color  than  the  inner  ones ;  the  former  forming  the  sap-wood,  and 
the  latter  the  heart-wood.  When  the  log  dries,  and  thus  tends 
to  contract,  it  will  be  found  that  scarcely  any  contraction  takes 
place  in  the  medullary  rays;  but  it  must  take  place  along 
the  line  of  least  resistance,  viz.,  along  the  annual  rings,  thus 
causing  radiating  cracks,  and  drawing  the  rays  nearer  together 
on  the  side  away  from  the  crack.  This  action  is  exhibited  in 


640 


APPLIED   MECHANICS. 


FIG.  241. 


Fig.  241,  where  a  log  is  shown  with  two  saw-cuts  at  right 
angles  to  each  other ;  when  this  log  becomes  dry,  the  four 
right  angles  all  becoming  acute 
through  the  shrinkage  of  the 
rings. 

If  the  log  be  cut  into  planks  by 
parallel  saw-cuts,  the  planks  will, 
after  drying,  assume  the  forms 
shown  in  Fig.  242,  as  is  pointed 
out  in  Anderson's  "  Strength  of 
Materials,"  from  which  these  two 
cuts  are  taken. 

This  internal  construction  of  a 
plank  has  an  important  influence 
upon  the  side  which  should  be  uppermost  when  it  is  used  for 

flooring ;  for,  if  the  heart  side  is  up- 
permost, there  will  be  a  liability  to 
having  layers  peel  off  as  the  wood 
dries  :  indeed,  boards  for  flooring 
should  be  so  cut  as  to  have  the  an- 
nual rings  at  right  angles  to  the 
side  of  the  plank.  Before  discuss- 
ing any  other  considerations  which 
affect  the  adaptability  of  timber  to 
use  in  construction,  we  will  con- 
sider the  question  of  its  strength. 
§  237.  Strength  of  Timber.  —  In  this  regard  we  must 
observe,  that,  whereas  the  strength  and  elasticity  and  other 
properties  of  iron  and  steel  vary  greatly  with  its  chemical  com- 
position and  the  treatment  it  has  received  during  its  manufac- 
ture, the  strength,  etc.,  of  timber  is  much  more  variable,  being 
seriously  affected  by  the  soil,  climate,  and  other  accidents  of  its 
growth,  its  seasoning,  and  other  circumstances  ;  and  that  over 
many  of  these  things  we  have  no  control :  hence  we'  must  not 


FIG.  242. 


STRENGTH  OF   TIMBER.  641 

expect  to  find  that  all  timber  that  goes  by  one  name  has  the 
same  strength,  and  we  shall  find  a  much  greater  variation  and 
irregularity  in  timber  than  in  iron.  The  experiments  that  have 
been  made  on  strength -and  elasticity  of  timber  may  be  divided 
into  the  following  classes  :  — 

i°.  Those  of  the  older  experimenters,  except  those  made 
on  full-size  columns  by  P.  S.  Girard,  and  published  in  1798. 
A  fair  representation  of  the  results  obtained  by  them,  all  of 
which  were  deduced  from  experiments  on  small  pieces,  is  to 
be  found  in  the  tables  given  in  Professor  Rankine's  books, 
" Applied  Mechanics,"  "Civil  Engineering,"  and  "Machinery 
and  Millwork." 

2°.  Tests  made  by  modern  experimenters  on  small  pieces. 
Such  tests  have  been  made  by  — 

(a)  Trautwine :  Engineers'  Pocket-Book. 

(^)   Hatfield  :  Transverse  Strains. 

(f)   Laslett :  Timber  and  Timber  Trees. 

(d)  Thurston  :  Materials  of  Construction. 

(e)  A  series  of  tests  on  small  samples  of  a  great  variety  of  American 

woods,  made   for   the  Census  Department,  and   recorded  in 
Executive  Document  No.  5,  48th  Congress,  ist  session. 

3°.  Tests  made  by  Capt.  T.  J.  Rodman,  U.S.A.,  the  results 
of  which  are  given  in  the  "Ordnance  Manual." 

4°.  All  tests  that  have  been  made  on  full-size  pieces. 

In  regard  to  tests  on  small  pieces,  such  as  have  commonly 
been  used  for  testing,  it  is  to  be  observed,  that,  while  a  great 
deal  of  interesting  information  may  be  derived  from  such  tests 
as  to  some  of  the  properties  of  the  timber  tested,  nevertheless, 
such  specimens  do  not  furnish  us  with  results  which  it  is  safe 
to  use  in  practical  cases  where  full-size  pieces  are  used.  Inas- 
much as  these  small  pieces  are  necessarily  much  more  perfect 
(otherwise  they  would  not  be  considered  fit  for  testing),  having 
less  defects,  such  as  knots,  shakes,  etc.,  than  the  full-size  pieces, 


642 


APPLIED   MECHANICS. 


they  have  also  a  far  greater  homogeneity.  They  also  season 
much  more  quickly  and  uniformly  than  full-size  pieces.  In 
making  this  statement,  I  am  only  urging  the  importance  of 
adopting  in  this  experimental  work  the  same  principle  that  the 
physicist  recognizes  in  all  his  work ;  viz.,  that  he  must  not 
apply  the  results  to  cases  where  the  conditions  are  essentially 
different  from  those  he  has  tested. 

Moreover,  it  will  be  seen  in  what  follows,  that,  whenever 
full-size  pieces  have  been  tested,  they  have  fallen  far  short  of 
the  strength  that  has  been  attributed  to  them  when  the  basis 
in  computing  their  strength  has  been  tests  on  small  pieces  ; 
and,  moreover,  the  irregularities  do  not  bear  the  same  propor- 
tion in  all  cases,  but  need  to  be  taken  account  of. 

The  results  of  the  first  class  of  experiments  named  in  the 
following  table  are  taken  from  Rankine's  "  Applied  Mechanics ;" 
and,  inasmuch  as  the  table  contains  also  the  strengths  of  some 
other  organic  fibres,  it  will  be  inserted  in  full.  The  student 
may  compare  these  constants  with  those  that  will  be  given 
later. 


Modulus 

Kind  of  Material. 

Tenacity 
or  Resist- 
ance to 
Tearing. 

Modulus  of 
Tensile 
Elasticity. 

Resist- 
ance to 
Crush- 
ing. 

Modulus 
of 
Rupture. 

Resist- 
ance to 
Shearing 
along 

of 

Shearing 
Elasticity 
along  the 

Grain. 

Grain. 

Ash  ....... 

I7OOO 

I6OOOOO 

QOOO 

(  12000 

1400 

76000 

I  I4OOO 

Bamboo     

6300 

- 

- 

- 

<-• 

- 

Beech    .     .  :it-.^\  $•,-_„/ 

II5OO 

1350000 

9360 

(     9000 
|  I2OOO 

\- 

- 

Birch     

I5OOO 

164^000 

6400 

II7OO 

_ 

Blue  gum  

8800 

(  16000 
(  2OOOO 

}- 

- 

Box        

2OOOO 

10300 

_ 

_ 

Bullet-tree      .... 

- 

14000 

(15900 
(  I6OOO 

I- 

ritof! 

Cedar  of  Lebanon  .     . 

II400 

486000 

5860 

7400 

- 

- 

STRENGTH  OF   TIMBER, 


643 


Kind  of  Material. 

Tenacity 
or  Resist- 
ance to 
Tearing. 

Modulus  of 
Tensile 
Elasticity. 

Resist- 
ance to 
Crush- 
ing. 

Modulus 
of 
Rupture. 

Resist- 
ance to 
Shearing 
along 
Grain. 

Modulus 
of 
Shearing 
Elasticity 
along  the 
Grain. 

(  IOOOO 

j 

Chestnut    

i     to 

>  II40000 

- 

10660 

- 

- 

(  I3OOO 

) 

Cowrie                      • 

_ 

_ 

IIOOO 

_ 

_ 

Ebony             .... 

19000 

27000 

(     700000 

I   6000 

f 

Elm            

14000 

to 

>  in^on 

<     to 

76000 

(  1840000 

) 

I  9700 

)       °° 

Fir,  Red  pine     .     .     . 

(  I2OOO 

)     to 

(  14000 

1460000 
to 
1900000 

5375 
to 
6200 

f    7100 

)  9540 

500 
800 

62OOO 
Il6ooo 

"     Yellow  pine  (Am.) 

- 

- 

5400 

- 

- 

- 

"     Spruce    .     ...~^.J 

12400 

(  1400000 
to 
(  1800000 

I  - 

(  9900 
1  12300 

|     600 

- 

"     Larch     .     .  V  f 

(     9000 
(  IOOOO 

900000 
to 
1360000 

|  5570 

(  5000 

(   IOOOO 

970 
1700 

|    - 

Hoxen  yarn    .... 

25000 

- 

- 

- 

- 

- 

Hazel    ....    w. 

16000 

- 

- 

- 

- 

- 

(  I2OOO 

~\ 

Hempen  rope     .     .    '. 

i     to 

V         - 

- 

_ 

- 

- 

(  I6OOO 

) 

Ox-hide,  undressed 

6300 

- 

- 

- 

- 

- 

Hornbeam      .... 

20000 

- 

- 

- 

- 

- 

Lancewood     .... 

23400 

- 

-     ' 

- 

- 

- 

Ox-leather      .... 

42OO 

24300 

- 

- 

- 

- 

Lignum-vitae  .... 

IISOO 

- 

9900 

I2OOO 

- 

- 

Locust       ..... 

l6oOO 

_ 

Mahoaany           ... 

(     8000 

<     to 

l> 

8200 

(  7600 

1    . 

(  21800 

) 

(  11500 

1 

Maple 

10600 

_ 

Oak,  British  .... 

;•  i* 

IOOOO 

(  IOOOO 

1  13600 

"     Dantzic      .    .     . 

- 

- 

7700 

8700 

(  IOOOO 

\ 

5-2300 

82OOO 

"     European  .     .    . 

)     to 

(19800 

f  I2OOOOO 

C  1750000 

}  - 

- 

"     American  red 

10250 

2I5OOOO 

6000 

10600 

J 

644 


APPLIED  MECHANICS. 


Resist- 

Modulus 

Kind  of  Material. 

Tenacity 
or  Resist- 
ance to 
Tearing. 

Modulus  of 
Tensile 
Elasticity. 

Resist- 
ance to 
Crush- 
ing. 

Modulus 
of 
Rupture. 

ance  to 
Shearing 
along 
Grain. 

of 
Shearing 
Elasticity 
along  the 

Grain. 

Silk  fibre  

52000 

1300000 

. 

. 

. 

. 

Sycamore  

13000 

1040000 

- 

9600 

- 

- 

) 

Teak,  Indian      .    .     . 

15000 

24OOOOO 

12000 

\  IQOOO 

1- 

— 

"       African     .     .     . 

21000 

23OOOOO 

- 

14980 

- 

- 

Whalebone     .... 

7700 

- 

- 

- 

- 

- 

Willow  

_ 

_ 

6600 

_ 

_ 

Yew  

8000 

In  regard  to  the  tests  of  the  second  class,  a  few  comments 
are  in  order  :  — 

i°.  These  experiments,  like  those  of  the  first  class,  were  all 
made  upon  small  pieces ;  and  the  results  are  correspondingly 
high. 

The  usual  size  of  the  specimens  for  crushing  being  one  or 
two  square  inches  in  section,  and  of  those  for  transverse 
strength  being  about  two  inches  square  in  section  and  four  or 
five  feet  span,  those  for  tension  had  even  a  much  smaller  sec- 
tion than  those  for  compression  ;  as  it  is  necessary,  in  order  to 
hold  the  wood  in  the  machine,  to  give  it  very  large  shoulders. 

The  only  exception  to  this  is  the  tests  of  Sir  Thomas  Las- 
lett,  an  account  of  which  is  given  in  his  "  Timber  and  Timber 
Trees,"  and  also  in  D.  K.  Clark's  "  Rules  and  Tables."  In  these 
tests  he  gives  very  much  lower  tensile  strengths  than  those 
given  above ;  and  he  states  that  his  specimens  were  three  inches 
square,  but  does  not  say  how  he  managed  to  hold  them  in  such 
a  way  as  to  subject  them  to  a  direct  tensile  stress.  His  results 
for  crushing  and  transverse  strength  are  about  as  great  as 


STRENGTH  OF   TIMBER.  645 

those  given  in  Rankine's  tables,  and  as  were  obtained  by  the 
other  experimenters  on  small  pieces,  as  his  specimens  were  of 
about  the  same  dimensions  as  those  used  by  the  others.  The 
figures  obtained  by  these  experimenters  will  only  be  given  inci- 
dentally, as  — 

(a)  They  are  very  similar  to  those  given  in  Rankine's  table. 

(b)  They  are  not  suitable  for  practical  use  on  the  large 
scale. 

(c)  While  they  have  been  used,  it  has  only  been  done  by 
employing  a  very  large  factor  of  safety  for  timber. 

The  series  of  tests  made  for  the  Census  Department,  and 
recorded  in  Executive  Document  No.  5,  48th  Congress,  first 
session,  form  a  very  interesting  series  of  experiments  upon 
small  specimens  of  an  exceedingly  large  number  of  American 
woods.  In  order  to  have  working  figures,  we  should  need  to 
test  large  pieces  of  the  same ;  as  the  proportion  between  the 
strengths  of  the  different  kinds  would  be  liable  to  be  different 
in  the  latter  case. 

The  only  record  of  Rodman's  experiments  available  is  a 
table  of  results  in  the  "  Ordnance  Manual."  These  are  lower, 
as  a  rule,  than  those  obtained  by  the  experimenters  of  the  first 
or  second  class.  This  is  to  be  accounted  for  by  the  fact  that, 
while  he  did  not  experiment  on  full-size  pieces,  he  used  much 
larger  pieces  than  those  heretofore  employed  ;  his  specimens 
for  transverse  strength,  many  of  which  are  still  stored  .at  the 
Watertown  Arsenal,  being  5f  inches  deep,  2$  inches  thick,  and 
5  feet  span. 

The  fourth  class  of  tests  are  those  which  furnish  reliable 
data  for  use  in  construction ;  and  we  will  proceed  to  a  consid- 
eration of  these,  taking  up  (i°)  tension,  (2°)  compression,  (3°) 
transverse  strength,  and  (4°)  shearing  along  the  grain. 


646  APPLIED  MECHANICS. 

TENSION. 

In  all  cases  where  the  attempt  has  been  made  to  experiment 
upon  the  tensile  strength  of  timber,  a  great  deal  of  difficulty 
has  been  encountered  in  regard  to  the  manner  of  holding  the 
specimens.  In  all  cases  it  has  been  found  necessary  to  pro- 
vide them  with  shoulders,  each  shoulder  being  five  or  six  times 
as  long  as  the  part  of  the  specimen  to  be  tested,  and  to  bring 
upon  these  shoulders  a  powerful  lateral  pressure,  to  prevent 
the  specimen  from  giving  way  by  shearing  along  the  grain,  and 
pulling  out  from  the  shoulder,  instead  of  tearing  apart. 

The  specimens  tested  have  generally  had  a  sectional  area 
less  than  one  square  inch,  and  it  seems  almost  impossible  to 
provide  the  means  of  holding  larger  specimens.  This  being 
the  case,  it  is  plain,  that,  whenever  timber  is  used  as  a  tie-bar 
in  construction  (except  in  exceedingly  rare  and  out-of-the- 
way  cases),  it  will  give  way  by  some  means  other  than  direct 
tension  ;  i.e.,  either  by  the  pulling-out  of  the  bolts  or  fastenings, 
and  the  consequent  shearing  of  the  timber,  or  else  by  bending 
if  there  is  a  transverse  stress  upon  the  piece ;  and,  this  being 
the  case,  these  other  resistances  should  be  computed,  instead 
of  the  direct  tension.  Hence,  while  the  direct  tensile  strength 
of  timber  may  be  an  interesting  subject  of  experiment,  it  can 
serve  hardly  any  purpose  in  construction  ;  and  the  conclusion 
follows,  that  the  resistances  of  timber  to  breaking  we  may 
expect  to  meet  in  practice  are  its  crushing,  transverse,  and 
shearing  strength.  Indeed,  the  use  of  timber  for  a  tie-bar 
should  be  avoided  whenever  it  is  possible  to  do  so  ;  and,  when 
it  is  used,  the  calculations  for  its  strength  should  be  based 
upon  the  pulling-out  of  the  fastenings,  the  shearing  or  splitting 
of  the  wood,  etc.,  and  not  on  the  tensile  resistance  of  the  solid 
piece. 

Moreover,  when  a  wooden  tie-bar  is  subjected  not  merely 
to  direct  tension,  but  also  to  a  bending-moment,  whether  the 
latter  is  caused  by  a  transverse  load,  or  by  an  eccentric  pull,  as 
it  generally  is  in  the  case  of  timber  joints,  we  must  compute 


TENSION.  647 


the  greatest  tension  per  square  inch  at  the  outside  fibre  due  to 
the  bending,  and  to  that  add  the  direct  tension  per  square 
inch :  and  this  sum  must  be  less  than  the  modulus  of  rupture 
if  the  piece  is  not  to  give  way;  i.e.,  the  modulus  of  rupture, 
and  not  the  ultimate  tensile  strength  per  square  inch,  must  be 
our  criterion  of  breaking  in  such  a  case,  the  working-strength 
per  square  inch  being  the  modulus  of  rupture  divided  by  a 
suitable  factor  of  safety. 

.   COMPRESSIVE    STRENGTH. 

Tests  of  the  compressive  strength  of  full-sized  wooden  col- 
umns are,  with  the  exception  of  one  set  of  tests,  of  very  recent 
date,  and  have  not  yet  found  their  way,  to  any  extent,  into  our 
text-books  and  engineers'  handbooks.  It  is  therefore  only 
suitable,  that,  before  enumerating  them,  we  should  observe 
what  formulae  have  been  given  in  these  books  for  the  computa- 
tion of  timber  columns  in  practice,  what  experimental  basis 
these  formulae  rest  upon,  and  how  they  coincide  with  the 
facts. 

The  oldest  formulae  are  those  of  Euler.  His  formulae  for 
the  strength  of  such  circular  columns  as  yield  by  bending  are 
as  follows  :  — 

For  ends  fixed  in  direction, 

(3.1416)3    </4 

A ~/29 

For  rounded  ends, 

(3.1416)3    a* 


where  P  =  breaking-weight  in  pounds,  and  E  =  modulus  of 
elasticity  of  the  material.  For  wood,  Weisbach  gives  for  use 
in  Euler's  formulae,  E  =  1664000,  and  crushing-strength  per 
square  inch  —  6770. 


648  APPLIED  MECHANICS. 

According  to  Hodgkinson,  we  should  take  one-ninth  of  the 
breaking-strength  of  a  cast-iron  pillar,  as  given  by  equations 
(i),  (2),  (3),  (4),  (5),  §  211  ;  this  being  the  rule  given  by  Mr. 
James  B.  Francis  in  his  book,  where  he  has  computed  and 
tabulated  the  strength  of  pillars  of  the  ordinary  sizes  used  in 
practice,  by  means  of  Mr.  Hodgkinson's  formulae. 

In  Gordon's  formula  the  constants  used  were  determined 
in  such  a  way  as  to  make  the  results  agree  as  nearly  as  possible 
with  Hodgkinson's  experiments.  The  formulae  devised  by  Gor- 
don himself  refer  only  to  cylindrical  and  hollow  cylindrical 
columns.  A  formula  devised  by  the  same  course  of  reasoning, 
and  also  depending  for  its  constants  on  Hodgkinson's  experi- 
ments, but  so  arranged  as  to  be  applicable  to  any  form  of 
section  whatever,  is  given  by  Professor  Rankine.  For  wood 
it  is  as  follows  : 


I  + 


where  P  —  breaking-strength  in  pounds,  S  =  sectional  area  in 
square  inches,  /  =  length  in  inches,  r  =  least  radius  of  gyra- 
tion in  inches. 

Besides  the  above,  we  have  formulae  which  are  practically 
Rankine' s  formulae  with  the  constants  changed.  One  of  these 
is  that  of  Mr.  C.  Shaler  Smith  as  given  in  Trautwine's  "  Pocket- 
Book,"  and  applicable,  as  he  claims,  to  a  square  or  rectangular 
column  of  white  or  yellow  pine.  It  is  as  follows : 


i  +  0. 


004^ 


where  P  =  breaking- weight  in  pounds,  6"  =  sectional  area  in 
square  inches,  /  =  length  in  inches,  d  =  least  side  of  rectangle. 
I  have  computed,  by  means  of  these  formulae,  the  breaking 
weights  of  certain  oak  columns,  with  the  following  results  :  — 


CO MP RE  SSI  VE   S  TRENG  TH. 


649 


Length  =14  feet. 


Diameter 
10.5  Inches. 

Diameter 
9.5  Inches. 

Euler      . 

(  Flat  ends  
£  Rounded  ends  .  .  . 
(  Flat  ends  .  .  .  .  -  . 

586214 

347M7 
4.6^261 

479858 
232623 
260164. 

Rankine 

(  Rounded  ends  .  .  . 
(  Flat  ends  .  ^  %  .  . 

263615 
420000 

191286 

•2  i  27CK 

Francis  . 

(  Rounded  ends  .  .  . 

140000 

O  x  •*•  tyj 
104265 

If  we  use  the  average  of  the  results  obtained  from  the  oak 
columns  tested  at  Watertown,  which  will  be  referred  to  later, 
we  should  find,  for  the  breaking-weights  of  the  above  columns 
with  flat  ends,  about  277000  and  227000  Ibs.  respectively. 

A  glance  at  the  above  results  will  show  that  they  differ  very 
much  from  each  other,  and  the  question  naturally  arises  as  to 
the  trustworthiness  of  the  experimental  data  on  which  they 
are  based. 

The  constants  used  in  Euler's  formulae  are  not  deduced 
from  any  experiment  on  the  breaking  of  a  column. 

Rankine's  and  Francis's  have,  for  experimental  basis,  the 
experiments  of  Hodgkinson.  He  made  a  very  large  number 
of  tests  of  cast-iron  columns,  none  of  which  were  as  large  as 
those  used  in  practice.  On  oak  columns,  he  made  seventeen 
experiments  on  as  many  columns,  all  cut  from  one  good  plank 
of  Dantzic  oak,  the  largest  of  which  was  five  feet  long  and  two 
inches  square.  Of  these  seventeen,  only  seven  were  used  in 
deducing  his  formulae. 

It  is  plain  that  such  data  are  insufficient,  and  cannot  furnish 
us  reliable  information  as  to  the  strength  of  a  column. 

As  to  Mr.  C.  Shaler  Smith's  formula,  it  is  based  upon  some 
experiments  made  by  himself,  which  have  never  been  published. 
The  student  can  easily  satisfy  himself  as  to  how  accurately  it 
represents  the  facts  on  the  large  scale,  by  computing  by  it  the 
strength  of  some  of  the  columns  tested,  of  which  an  account 


650  APPLIED   MECHANICS. 

will  shortly  be  given.  Trautwine,  in  his  "  Handbook,"  states 
that  it  is  "for  the  breaking-loads  of  either  square  or  rectangular 
pillars  or  posts,  of  moderately  seasoned  white  and  common 
yellow  pine,  with  flat  ends,  firmly  fixed,  and  equally  loaded." 

Mr.  Smith,  as  I  understand,  however,  intended  it  to  meet 
the  case  of  such  ill-fitting  joints  as  occur  in  practice,  and  not 
for  perfectly  even  bearings. 

TESTS   OF   FULL-SIZE   COLUMNS. 

The  only  tests  of  full-size  columns  of  which  I  have  any 
knowledge  are  :  — 

i°.  Trautwine,  in  his  "Handbook,"  speaks  of  some  tests  of 
wooden  pillars  20  feet  long  and  13  inches  square,  made  by 
David  Kirkaldy,  which,  as  he  says,  gave  results  agreeing  with 
Mr.  C.  Shaler  Smith's  rule. 

2°.  A  series  of  tests  made  at  the  Watertown  Arsenal  for 
the  Boston  Manufacturers'  Mutual  Fire  Insurance  Company, 
under  the  direction  of  the  author. 

3°.  Eleven  tests  of  old  spruce  pillars  made  at  the  Water- 
town  Arsenal,  for  the  Jackson  Company,  under  the  direction 
of  Mr.  J.  R.  Freeman,  and  reported  in  the  Journal  of  the 
Assoc.  Eng.  Societies  for  November,  1889. 

4°.  The  tests  that  have  been  made  at  the  Watertown  Arse- 
nal on  the  government  testing-machine. 

5°.  Besides  these,  the  writer  has  very  recently  had  his 
attention  called  to  a  series  of  tests  of  full-size  columns  of  oak 
and  fir,  made  by  P.  S.  Girard  in  1798,  which  give  results  agree- 
ing very  well  with  the  modern  results  on  full-size  columns. 
Why  these  tests  should  have  been  lost  sight  of,  and  Hodgkin- 
son's  always  used  instead,  is  incomprehensible. 

In  regard  to  the  first,  no  details  or  results  are  given  :  hence 
nothing  will  be  said  about  them. 

In  regard  to  the  second,  a  summary  only  will  be  presented 
here,  the  reader  being  referred  for  details  to  my  published 
report  entitled  "  Strength  of  Wooden  Columns." 


TESTS  OF   YELLOW-PINE  POSTS  AND  BLOCKS.          651 


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652 


APPLIED   MECHANICS. 


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TESTS  OF  OLD   AND  SEASONED    WHITE-OAK  POSTS.    653 


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654 


APPLIED   MECHANICS. 


In  all  the  experiments  enumerated  in  the  tables  given 
above,  the  columns  gave  way  by  direct  crushing,  and  hence 
the  strength  of  columns  of  these  ratios  of  length  to  diameter 
can  properly  be  found  by  multiplying  the  crushing-strength  per 
square  inch  of  the  wood  by  the  area  of  the  section  in  square 
inches. 

This  conclusion  is  deduced  from  the  fact  that  the  deflections 
were  measured  in  every  case,  and  found  to  be  so  small  as  not  to 
exert  any  appreciable  effect. 

In  regard  to  other  tests  of  this  same  set,  there  were  eight 
tests  made,  in  addition  to  those  already  enumerated ;  and  of 
these,  five  were  loaded  off  centre.  A  summary  of  the  results  is 
appended,  together  with  a  comparison  of  their  actual  strength 
with  that  which  would  be  computed  on  the  basis  of  4400  per 
square  inch  for  yellow  pine,  and  3000  for  oak. 


Weight, 
in  Ibs. 

Length,  in 
feet  and 
inches. 

Diameter 
of 
Column. 

Diam- 
eter of 
Core. 

Sectional 
Area,  in 
square 
inches. 

Eccen- 
tricity, 
in 
inches. 

Ultimate 
Strength. 

Computed 
Ultimate 
Strength. 

ft.          in. 

2,  2d  series 

320 

II     11.27 

9.92 

i-53 

7545 

2-33 

265000 

331980 

5,  3d  series 

298 

12       6.8 

$    8f>? 

- 

63.1 

2.07 

24OOOO 

277640 

i,  3d  series 

386 

12       9.3 

(   8-75) 
I   8.92! 

- 

76.04 

2.25 

280000 

334576 

i,  2d  series 

451 

II     1  1  .4 

10.95 

i.  80 

92.16 

2.75 

170000 

276480 

3,  2d  series 

236 

II     II.  2 

8.2 

-.55 

50.92 

,.9I 

IOOOOO 

152760 

These  results  exhibit  a  great  falling-off  of  strength  due  to 
the  eccentricity  of  the  load ;  and  if  we  observe,  that,  whenever 
the  beam  on  one  side  of  a  column  is  loaded  differently  from 
that  on  the  other  side,  we  have  an  eccentric  loading,  and  hence 


STRENGTH  OF   TIMBER.  655 

a  falling-off  in  strength,  we  must  conclude  that  this  should  be 
taken  into  account.  Probably  the  best  way  to  proceed  in  the 
matter  is  to  compute  always  the  greatest  eccentricity  possible 
in  any  given  case,  and  to  compute  from  that  the  additional 
stress  on  the  column  due  to  the  bending  consequent  on  this 
eccentricity  by  the  principles  of  the  short  strut,  already  ex- 
plained in  §  207.  This  will  always  be  on  the  safe  side.  The 
three  remaining  experiments  of  the  set  were,  (i°)  Two  tests  of 
whitewood  columns :  these  gave  a  crushing-strength  of  3009 
Ibs.  per  square  inch.  The  columns  were,  however,  brittle,  and 
did  not  give  warning  of  fracture.  (2°)  One  yellow-pine  square 
column  of  a  sectional  area  of  68.8  square  inches,  and  length  12 
feet  6.85  inches,  tested  by  resting  one  end  on  a  thick  yellow- 
pine  bolster,  crushing  this  bolster  at  right  angles  to  the  grain. 

Maximum  load  on  the  post  while  the  bolster  was  in,  was 
120000  Ibs.  =  1744  Ibs.  per  square  inch.  Under  this  load  a 
crack  at  end  of  post  was  enlarged,  giving  evidence  that  failure 
of  the  post  was  imminent.  The  bolster  in  the  mean  time  had 
become  thoroughly  cracked,  owing  to  the  unequal  distribution  of 
the  load  on  the  bolster,  from  imperfect  workmanship.  Slight 
cracks  followed  the  first  snapping  sounds  heard  at  20000  Ibs. 
compression.  The  cracks  gradually  developed  as  the  loads  were 
increased,  the  side  nearest  the  heart  of  the  bolster  sliding  off. 

The  post  was  taken  from  machine  and  bolster  removed. 
The  post  was  cut  off  i  J  inches  at  the  end,  and  squared ;  the 
total  length,  after  cutting,  being  149.35  inches.  Ultimate 
strength,  375000  Ibs.  =  5451  Ibs.  per  square  inch. 

A  few  of  the  conclusions  in  regard  to  these  sizes  of  posts  will 
now  be  quoted  from  the  report  of  these  tests,  in  which  some 
general  recommendations  are  made,  and  others  having  special 
reference  to  mill  columns  :  — 

i .  I  should  recommend  that  the  longitudinal  holes  in  wooden  mill 
columns  be  bored  from  one  end  only,  and  that  all  posts  be  rejected  in 


656  APPLIED  MECHANICS. 

which  the  eccentricity  at  the  other  end  is  greater  than  a  given  small 
amount,  as  three-quarters  of  an  inch.  This  recommendation  is  made 
in  view  of  the  fact  that  holes  bored  from  the  two  ends  are  very  liable 
not  to  meet  in  the  middle. 

2.  I  should  recommend  that  mill  columns  be  not  tapered,  as  the 
tapering  is  a  source  of  weakness. 

3.  I  should  also  recommend  that  square  columns  be  used  in  mills, 
instead  of  round  ones,  for  the  reason  that  the  timber  comes  to  the 
wharf  in  the  form  of  square  logs,  and,  when  the  columns  are  made 
round,  they  are  cut  from  the  square  form  ;  and  this  cutting-away  of 
the  wood  is  so  much  loss  of  strength. 

4.  The  strength  of  a  column  of  hard  pine  or  oak,  with  "  flat  ends," 
the  load  being  uniformly  distributed  over  the  ends,  and  of  the  diame- 
ters tested,  is  practically  independent  of  the  length,  for  the  ratios 
of  length  to  diameter  used  in  the  tests,  such  columns  giving  way 
practically  by  direct  crushing  :   the  deflection,  if  any,  being  as  a 
rule  very  small,  and  exerting  no  appreciable  influence  on  the  break- 
ing-strength. 

5.  The  crushing-strength  per  square  inch  varies  considerably  in 
specimens  of  different  degrees  of  seasoning,  also  in  large  and  small 
s-pecimens.    The  average  crushing-strength  of  such  yellow-pine  posts 
as  were  tested,  not  thoroughly  seasoned,  and  not  very  green,  is  about 
4400  Ibs.;   whereas  for  such  oak  as  was  furnished  me,  which   was 
green  and  knotty,  but  no  more  so  than  is  usual  for  use  in  building, 
the  average  is  about  3200  Ibs. 

6.  I  would  recommend  the  use  of  iron  caps  and  pintles,  instead  of 
wooden  bolsters,  as  wood  is  very  weak  to  resist  crushing  across  the 
grain,  and  the  wooden  bolster  will  fail  at  a  pressure  far  below  that 
which  the  column  is  capable  of  resisting  ;  and  the  unevenness  of 
the  pressure  brought  about  by  the  bolster  is  so  great  as  to  sometimes 
crack  the  column  at  a  pressure  far  below  what  it  would  otherwise 
sustain. 

7.  Any    cause    which    operates    to    distribute    the    pressure    on 
the   ends   unevenly,  or    to   force    its    resultant   out  of    centre,  is  a 
source  of  weakness,  and  brings  about  a  very  considerable  deflec- 


STRENGTH  OF  TIMBER.  657 

tion,  which  exerts  an  important  influence  in  reducing  the  breaking- 
strength. 

The  table  of  results  of  the  tests  on  old  and  seasoned  oak 
columns  were  made  upon  columns  that  had  been  in  use  for  a 
number  of  years  in  different  mills,  from  which  they  were  re- 
moved, and  replaced  by  new  ones.  Ten  of  them  had  been  in 
use  about  twenty-five  years,  and  the  remainder  for  shorter 
periods.  An  inspection  of  this  table  will,  I  think,  convince  the 
reader  that  it  would  not  be  safe  to  calculate  upon  a  higher 
breaking-strength  per  square  inch  in  these  than  in  the  green 
ones. 

TESTS    FOR    THE   JACKSON    COMPANY. 

Eleven  tests  of  old  spruce  pillars,  which  had  been  in  use 
in  a  cotton-mill  of  the  Jackson  Company,  were  tested  on  the 
government  machine  at  Watertown,  under  the  direction  of 
Mr.  J.  R.  Freeman.  The  manner  of  making  them  was  as  fol- 
lows: 

In  the  first  two  the  ends  were  brought  to  an  even  bear- 
ing. 

In  the  third  the  ends  came  to  an  even  bearing  under  a  load 
of  60000  pounds. 

In  the  fourth,  fifth,  ninth,  tenth,  and  eleventh,  the  cap, 
and  also  the  base-plate,  were  planed  off  on  the  back  to  a  slope 
of  i  in  24,  and  placed  with  their  inclinations  opposite. 

In  the  eighth  they  had  their  inclinations  the  same  way  one 
as  the  other. 

In  the  sixth  and  seventh  the  base-plate  was  not  used,  the 
larger  end  of  the  post  having  a  full  bearing  on  the  platform  of 
the  machine. 

The  results  are  given  in  the  following  table : 


658 


APPLIED  MECHANICS. 


Length 
in  Feet  and 
Inches. 

Diameter 
at 
Small  End, 
Inches. 

Diameter 
at 
Large  End, 
Inches. 

Area  at 
Small  End, 
Sq.  In. 

Ultimate 
Strength, 
Lbs. 

Ultimate 
Strength  per 
Sq.  In., 
Lbs. 

ft.        in. 

I 

10     4-75 

5.82 

7.78 

31.87 

142000 

4088 

2 

10     4. 

5-85 

7-49 

27-I5 

192800 

6225 

3 

10    5-75 

.    5.85 

7-74 

32.17 

I66IOO 

4900 

4 

10    5-5 

5-70 

7-77 

31.67 

108200 

5 

10    5.1 

5-70  • 

7-J70 

30.78 

105000 

6 

10    5.2 

5.80 

7.61 

30.39 

168000 

7 

9    7 

5-74 

7.81 

32.88 

194100 

8 

10    5.4 

5-85 

7.90 

34.21 

155000 

9 

10    4.9 

5-82 

7-77 

40.72 

96100 

10 

10    5-13 

5-73 

7-78 

125000 

ii 

•  10    4.38 

5-74 

7.81 

60000 

All  but  the  first  three  of  the  tests  were  made  with  inclined 
bearings  of  one  kind  or  another,  hence  the  ultimate  strength 
per  square  inch  is  only  given  here  for  the  first  three ;  which,  as 
Mr.  Freeman  says,  were  of  "  well-seasoned  spruce,  of  excellent 
quality."  Hence  the  average  crushing-strength  of  spruce  is 
doubtless  considerably  lower  than  the  average  of  these  three. 


TESTS    MADE    ON    THE    GOVERNMENT    MACHINE. 

In  Executive  Document  12,  4/th  Congress,  first  session, 
will  be  found  a  series  of  tests  of  white  and  yellow  pine  posts 
made  at  the  Watertown  Arsenal ;  and  these  tests  probably  fur- 
nish us  the  best  information  that  we  possess  in  regard  to  the 
strength  of  wooden  columns. 

The  summary  of  results  is  appended  :  — 


COMPRESSION  OF   WHITE-PINE  POSTS. 


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APPLIED  MECHANICS. 


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COMPRESSION  OF   WHITE-PINE  POSTS. 


661 


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662 


APPLIED   MECHANICS. 


COMPRESSION    OF    WHITE    PINE.  —  SINGLE  STICKS  AND  BUILT  POSTS. 
In  the  multiple  ones,  dimensions  of  each  stick  are  given. 


No. 
of 
Test. 

Weight. 

Average  Rate  of 
Growth. 
Rings  per  Inch. 

Dimensions  of  Post. 

Sectional 
Area. 

>SF 

J    j 

8-9fiJS 

$*h 

Ultimate  Strength. 

! 

1 

jA 

1 

"rt 
| 

Id 

i 

Ibs. 

in. 

in. 

in. 

sq.  in. 

in. 

Ibs. 

664 

153 

ii 

i77'So 

4.48 

11.65 

52.2 

0.0545 

IIOOOO 

2107 

665 

143 

10 

180.00 

4.48 

11.64 

52.1 

O.IOIO 

81500 

1564 

666 

I63 

5 

179.97 

4-47 

11.63 

52.0 

0.0895 

70000 

1346 

667 

228 

13 

180.00 

5-40 

11.30 

61.0 

0.0505 

160000 

2623 

668 

193 

5 

179-93 

5-6i 

">  73 

65.8 

0.0622 

156300 

2375 

669 

253 

5 

180.00 

5-64 

11.76 

66.3 

0.0608 

152300 

2297 

638 

fiK 

<i34 

315 

\    6 

i  s 

180.00 
180.00 

4-50 
4-50 

ii.  60 
"•59 

52-2 
52-2 

104.4 

(  0.0670  ) 

(  0.0750  j 

200000 

1916 

639 

(  182 
[167 

349 

7 

5 

180.00 
180.00 

4-52 
4.49 

11.66 
11.62 

52-7 
52.2 

104.9 

0.0645  1 

0.0670  J 

212000 

202  1 

640 

131 

lisa 

289 

7 
5 

180.00 
180.00 

4-53 
4-52 

n-59 
11-59 

52.5 
52-5 

105.0 

0.1060  | 
0.0955  J 

149000 

1419 

642 

$192 
1  306 

498 

6 
7 

179.98 
179.98 

5-57 
S.58 

n.  61 
ii.  61 

64.7 
64.7 

129.4 

0.0770) 
0.0390) 

215000 

1661 

643 

(  203 
j  189 

392 

17 
5 

179.92 
179.92 

5-65 
5-6i 

ii.  61 
11.62 

65.6 
65.2 

130.8 

0.0440 
0.0600 

261000 

1995 

644 

227 

<236 

463 

7 
7 

179-96 
179.96 

5.60 
5-60 

11.63 
11.62 

65.1 
65.1 

130.2 

0.0596 
0.0690 

257800 

1980 

648 

J206 

I  203 

409 

12 

10 

180.00 
180.00 

5-60 
5-6i 

11.72 
11.72 

65.6 
65.6 

131-2 

0.0590  | 
0.0700  ) 

268000 

2042 

649 

ISJi'* 

9 
4 

180.00 
180.00 

5.60 
5-61 

11.71 
11.74 

65.6 

65-9 

131-5 

0.0600 
0.0705 

277000 

2107 

650 

194 

I  190 

384 

12 

9 

180.00 
180.00 

5.61 
5-61 

ii-75 
11.71 

g? 

131.6 

0.0530 
0.0885 

240000 

1824 

645 

ISil* 

5 
7 

179.97 
179.97 

5-59 
5-6o 

11.59 
1  1.  60 

&!'•»•' 

0.0560  ) 
0.0540  J 

263200 

2028 

646 

i£i- 

6 

12 

180.00 
180.00 

5-59 
5-60 

ii.  61 
11.62 

64.9 
65.1 

130.0 

0.0493 
0.0620 

24QOOO 

1915 

647 

(  192 
\  192 

384 

II 
13 

180.00 
180.00 

5.62 
5-62 

11.62 
11.62 

65-3 
65-3 

130.6 

0.0630  j 
0.0700  ) 

248000 

1899 

678 

f  251 

/2I5 

466 

16 
8 

179.94 
179.94 

5.58 
5-57 

11.47 
11-45 

64.0 
63.8 

127.8 

0.0529 
0.0642 

245500 

1921 

679 

1  209 
1  190 

399 

12 

7 

180.00 
180.00 

5.62 
5.62 

11.76 
11.72 

66.1 
65-9 

132.0 

|  0.0664  \ 
1  0.0705  j 

249000 

1886 

680 

268 

J278 

546 

8 
8 

180.00 
180.00 

5.60 
5.61 

11.72 
"•73 

65.6 
65.8 

131.4 

0.0650  J 
0.0495  j 

278000 

2116 

663 

j  230 

J208 

438 

\M 

<    6 

180.00 
180.00 

5.60 
5-63 

ii«75 
"•75 

65.8 
66.2 

132.0 

0.0621  ) 
0.0657  j 

300000 

2273 

676 

IS 

385 

12 

6 

179.94 
179.94 

5.60 
5.61 

11.71 
11-73 

65-6 
65.8 

i3i-4 

j  0.0530  | 
'  0.0593  J 

274500 

2089 

677 

IS!"8 

6 
6 

180.00 
180.00 

5-6i 
5-68 

11.72 
11.72 

65.7 
65-4 

131-1 

0-0551 
0.0625 

255000 

1945 

COMPRESSION  OF   WHITE   PINE. 


663 


COMPRESSION    OF    WHITE    PINE.—  Concluded. 
SINGLE  STICKS  AND  BUILT  POSTS. 


No. 
of 
Test. 

Weight. 

i 

<5   £ 

Dimensions  of  Post. 

Sectional 
Area. 

1  r»H«n 

Ultimate  Strength. 

i 

1 

i 

1 

le 

Ibs. 

in. 

in. 

in. 

sq.  in. 

in. 

Ibs. 

(175 

18 

180.00 

4-52 

ii  .62 

52-5) 

(  0.0460) 

690 

226 

600 

ii 

180.00 

5-56 

11.70 

65.0^169.3 

<  0.0580^ 

310000 

1831 

18 

180.00 

4.46 

11.62 

51-8) 

(  0.0480) 

691 

(3 

520 

9 
M 

179.98 
I79-98 

4.48 
5-56 

n.  60 
1  1.  60 

52.0) 

64.  5  [168.2 

(0.0526  ) 
]  0.0430  | 

372500 

2215 

(i59 

12 

I79-98 

4-45 

n.  61 

51-7) 

(  0.0390  ) 

(248 

(13 

I77-25 

5-62 

ii.  60 

65.2) 

(  0.0580  ) 

692 

153 

614 

12 

177-25 

4-5° 

1  1.  60 

52.2  \  i8a-.2 

{0.0641  \ 

363000 

1992 

(213 

(    5 

177-25 

5-6o 

J1-57 

64.8) 

(  0.0768  ) 

(  T5Z 

ii 

180.00 

4-5° 

ii.  60 

52.2! 

(  0.0460) 

687 

]  218 

536 

8 

180.00 

5-58 

11.62 

64.  8  [  169.4 

jo.  0587  [ 

325500 

1919 

(  167 

9 

180.00 

4-52 

n-59 

52.4) 

(0.0533) 

688 

(176 
236 

575 

10 
IO 

180.00 
i  80  .  oo 

4-50 
4.62 

1  1.  60 

ii  .60 

52.2) 
6S'2\  169  6 

(  0.0565  ) 
{  0.0645  } 

306000 

1804 

(163 

II 

180.00 

4-50 

1  1.  60 

52-2) 

(  0.0703  ) 

(188 

(    7 

180.05 

4.48 

ii.  60 

52-0) 

(  0.0510  ) 

689 

U3 

(  159 

55° 

ii 
9 

180.05 
180.05 

5.60 
4.46 

ii  .62 
n-57 

65.1     168.7 
51-6) 

]o.o66o[ 
(  0.0789  ) 

340000 

2015 

681 

j  192 

')  146 

U57, 

>6BB 

f 

\.  9 

179-95 

179-95 
179-95 

4-47 
4-52 
4.48 

ii.  60 
11.65 
11.65 
11.65 

52-5J 

("0.0700] 
^  0.0714  ^ 

[o.  0762] 

362000 

1734 

("161" 

f    7 

180.00 

4-5° 

11.63 

52-3! 

{0.0612') 

682 

1*63 

r 

180.00 
180.00 

4-50 
4-49 

ii  .64 
1  1.  60 

0.0546  ! 
0.0542  f 

414000 

1980 

1.  15I> 

I   9 

180.00 

4-50 

ii  .63 

52-3] 

0.0916  J 

[169' 

f  10 

180.03 

4.46 

11.63 

("0.0570] 

683 

j  219 
]2i8 

>787 

1  9 

180.03 
180.03 

5-62 
5-6o. 

11.69 
ii  .70 

234-9 

J  0.0530  ! 
]  0.0494  ( 

501000 

2133 

Ii8i, 

t    7 

180.03 

4  46 

11.61 

, 

lo.osgoj 

{146" 

fio 

180.00 

4-5° 

11.64 

52-4] 

[0.0664] 

684 

2IO 
234 

756 

} 

180.00 
180.00 

5*60 

n-59 
11.58 

j  0.0610  1 
!  0.0548  f 

529000 

2255 

,l66. 

I  10 

180.00 

4.48 

11.61 

52.0] 

1.0.05I4J 

CHS' 

fio 

180.00 

4-50 

1  1.  60 

52.2] 

{0.0645  1 

685 

1  209 

lio 

1  12 

180.00 
180.00 

5-6i 

11.62 
ii  .62 

65^2  f234-8 

0.0650  1 
0.0506  f 

430000 

1831 

U66, 

I  9 

180.00 

4-48 

11.62 

52.  oj 

0.0546J 

[M5 

I 

fn 

180.00 

4.48 

ii.  61 

52.0! 

fo.osoo] 

686 

1 

^37 

10 

1  I2 

180.00 
180.00 

4-5° 
4-50 

11.56 
11-36 

j  0.0315  1 
]  0.0543  [ 

395ooo 

1903 

LISO 

1 

I   8 

180.00 

4-52 

ii.  61 

52-5J 

{  0.0680  j 

664 


APPLIED   MECHANICS. 


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COMPRESSION  OF   YELLOW-PINE  POSTS. 


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COMPRESSION  OF   YELLOW-PINE  POSTS. 


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APPLIED  MECHANICS. 


COMPRESSION    OF    YELLOW    PINE. 
SINGLE  STICKS  AND  BUILT  POSTS. 


No. 
of 
Test. 

Weight, 
in  bs. 

Average  Rate  of  1 
Growth. 
Rings  per  Inch.  | 

Dimensions  of  Post. 

Sectional 
Area. 

i! 

Ultimate  Strength. 

! 

1 

"1 

1 

M 

& 

in. 

in. 

in. 

sq.  in. 

673 

260     ) 

1180.05 

4.09 

"•35 

46.04    ) 

(  142200 

3065 

6730 

13 

ii 

20.00 

4.01 

4.01 

16.08 

0.0370 

<   94000 

5846 

7*   ) 

20.00 

3-95 

3-97 

15-68   ) 

(   99600 

6352 

674 

233 

17 

iSo.OO 

4-Si 

ii.  60 

52-30 

0.0492 

131500 

2515 

675 

194 

22 

iSo.OO 

4-34 

ii.  60 

50-30 

0.0490 

I2I200 

2410 

490 

269 

- 

iSo.OO 

5-05 

12.10 

61.10 

- 

230000 

3764 

670 

309 

12 

iSo.OO 

5-65 

11.74 

66.30 

0-0455 

205900 

3106 

489 

.  287 

- 

180.08 

5.85 

12.05 

70.50 

- 

250000 

3546 

654 

1326 
^290 

616 

[II 

1    6 

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iSo.OO 

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5-62 

11.71 
II.7I 

§:Sl-3'.8 

0.0418  ) 
0.0540  i 

470000 

3566 

655 

(292 
I  329 

621 

Hi 

179-93 
179-93 

5-64 

11.72 
11.72 

S:;  }  '32.2 

0.0292 
0.0315 

SSOOOO 

4387 

656 

J282 

569 

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iSo.OO 

5-6i 

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II.7I 

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j  0.0466  f 

480000 

3653 

651 

j242 

537 

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iSo.OO 
iSo.OO 

5-58 
5.58 

11.71 
11.71 

ell  h0-6 

0.0620  ) 
0.0514  J 

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2756 

652 

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680 

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{16 

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5.58 

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0.0395  1 
0.0360  ) 

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4506 

653 

IIS 

621 

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\  I2 

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iSo.OO 

5-63 
5-59 

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11.68 

6^1^-2 

0-0559  } 
0.0550) 

436600 

3328 

657 

its 

659 

U 

179.96 
179.96 

5.63 
5-64 

11.72 

II.  7! 

66.-oi'3-° 

0-0375 
0.0305 

580000 

4394 

658 

J332 

65X 

1" 

179.98 
179.98 

5-59 
5-59 

11.71 
11.72 

Si!131'0 

0.0320  ) 
0.0436  ) 

448000 

3420 

659 

iS 

613 

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iSo.OO 
iSo.OO 

5.61 
5.61 

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"•73 

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0.0312 
0.0372 

600000 

4559 

660 

J294 

577 

13 

180.03 
180.03 

5.61 

5-63 

11.22 

11.24 

Si  i126-2 

0.0325 
0.0400 

5IOOOO 

4041 

661 

J274 

584 

i12 

1    9 

iSo.OO 
iSo.OO 

5-66 
5-60 

11.70 

11.72 

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0.0410  ) 
0.0365  \ 

410000 

3"i 

662 

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1*8 

iSo.OO 
iSo.OO 

5.61 

"•75 
"•75 

Is:'!131-8 

0.0540  ) 
0.0500  ) 

388000 

2944 

693 

(242 
276 

774 

{;; 

179.97 
179.97 

5-50 

ii.  61 
11.56 

52.2) 

63.6ji68.o 

0.0320 
0-0325 

564000 

3357 

(256 

(  23 

179.97 

4-49 

11.62 

52-2) 

0.0148 

694 

f«93 

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24 

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3027 

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0.0610 

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290 
(246 

743 

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16 

16 

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iSo.OO 
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59.0  >  161.1 
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0.0429 
0.0290 
0.0410 

474000 

2942 

[                       1 

COMPRESSION  OF   YELLOW  PINE. 


669 


COMPRESSION    OF    YELLOW    PINE.—  Concluded. 
SINGLE  STICKS  AND  BUILT  POSTS. 


•8       . 

•S"S  * 

3     t 

Dimensions  of  Post. 

>3 

Ultimate  Strength. 

No. 
of 
Test. 

Weight, 
in  Ibs. 

•I* 

Sectional 
Area. 

§     £  ji 

•5 

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| 

J 

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1 

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696 

217) 
253  j  660 
190) 

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(15 

180.00 
180.00 
180.00 

4-Si 
5'5o 
4.48 

11.24 
11.23 
11.23 

50-7) 
61.8  [162.  8 
50-3) 

!  0-0337  ) 
0.0567  > 
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480000 

2948 

697 

242) 

249  \  706 
215) 

('5 
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180.00 
180.00 
180.00 

4-52 
5-43 
4-47 

n.  60 
1  1.  60 
11.58 

52.4) 
63.0  >  167.2 

10.0240  1 
0.0330  J 
0.0336  ) 

540000 

3230 

698 

(224) 
j  255  [  775 

1 

179.94 
179.94 

4.46 
5-53 

n.  60 
11.70 

64  '.7  f  168.6 

(  0.0290) 
<  0.0385  > 

544000 

3227 

(296) 

(IS 

179.94 

4-50 

"•59 

52.2) 

(  0.0341  ) 

488 

911 

~ 

180.20 

J6.88 

15-75 
15-75 

:^i- 

- 

700000 

3268 

These  sets  of  tests  furnish  us  practically  the  only  reliable 
information  we  have  in  regard  to  the  strength  of  full-size  col- 
umns of  white  pine,  yellow  pine,  and  oak.  In  regard  to  white 
and  yellow  pine,  Mr.  Edward  F.  Ely,  instructor  in  architecture 
at  the  Massachusetts  Institute  of  Technology,  has  plotted  the 
average,  and  also  the  lowest,  results  of  the  tables ;  and,  from  an 
inspection  of  the  diagrams,  he  gives  the  following  rules  for 
determining  the  breaking-strength  of  a  column  with  flat  ends> 
the  load  being  evenly  distributed  over  the  ends  :  — 

Let  A  —  area  of  section  in  square  inches. 

f  =  constant  whose  value  is  given  in  the  tables   fol- 
lowing. 


Then 


-  =  ratio  of  length  to  least  side  of  rectangle,  all  the 
tests  having  been  made  on  rectangular  sections. 


Breaking-strength  =fA, 


APPLIED   MECHANICS. 


where  f  has  the  following  values  :  — 


White  Pine. 

Yellow  Pine. 

r 

/ 

r 

/m 

o  to  10 

2500 

o  to  15 

4000 

10  to  35 

2000 

15  to  30 

35°° 

35  to  45 

I5OO 

30  to  40 

3000 

45  to  60 

1000 

40  to  45 

2500 

45  to  5° 

2OOO 

50  to  60 

1500 

In  the  case  of  oak,  if  it  is  desired  to  apply  the  results  to 
greater  ratios  of  length  to  diameter  than  those  tested,  a  similar 
reduction  can  be  made  in  the  value  of  f  to  that  which  takes 
place  here  in  the  case  of  white  and  yellow  pine. 

For  the  ratios  of  length  to  diameter  tested,  f  =  3000  would 
seem  to  be  a  fair  value  to  use. 

§  238.  Factor  of  Safety. — Whereas,  in  the  case  of  iron 
bridge-work,  it  is  very  common  to  use  a  factor  of  safety  4,  the 
apparent  factors  of  safety  that  have  been  used  and  recom- 
mended for  timber  have  varied  very  greatly,  and  naturally  so, 
because  the  values  assumed  for  breaking-strength  have  been  so 
very  variable,  and  while  some  have  advised  the  use  of  apparent 
factors  of  safety  greater  than  4,  nevertheless  most  of  the  build- 
ing laws  only  require  an  apparent  factor  of  safety  3,  while 
making  use  of  values  of  breaking-strength  deduced  from  tests 
of  small  pieces. 

In  view  of  the  above  facts,  it  is  true  that  the  values  of  work- 
ing-strength used  in  many  cases  have  been  very  near  the  actual 
breaking-strength ;  and,  indeed,  it  would  be  impossible  to 
recommend  any  suitable  factor  of  safety  to  be  used  with  re- 


TRANSVERSE  STRENGTH  OF  TIMBER.  6/1 

suits  derived  from  tests  of  small  pieces.  But  if  the  true  values 
of  the  breaking-strength  as  derived  from  tests  of  full-size  pieces 
be  used,  it  would  seem  to  the  writer  that  a  factor  of  safety  4 
will  be  sufficient  for  most  ordinary  timber  constructions ;  i.e., 
that  we  should  use  for  working-strength  per  square  inch  one- 
fourth  of  the  breaking-strength  per  square  inch.  In  the  case 
of  mill-work,  and  in  other  cases  where  there  is  the  jarring  of 
moving  machinery,  it  is  advisable  to  use  a  somewhat  larger 
factor.  This  same  reasoning  will  also  apply  to  the  case  of 
beams  bearing  a  transverse  load,  where  they  are  designed  with 
reference  to  their  breaking-weight. 

§  239.  Transverse  Strength  of  Timber. — In  this  regard, 
the  common  theory  of  beams  has  already  been  explained  in 
§  185  et  seq. 

The  table  of  Rankine,  already  given,  represents  the  values 
of  modulus  of  rupture  that  have  been  in  common  use.  Other 
values,  not  differing  essentially  from  these,  are  given  by  Hat- 
field,  Laslett,  Thurston,  Trautwine,  and  others,  all  based  upon 
tests  of  small  pieces.  Confining  ourselves  to  tests  of  full-size 
pieces,  we  find  an  account  of  a  set  of  tests  attributed  by  D.  K. 
Clark,  in  his  "  Rules  and  Tables,"  to  Edwin  Clark  and  C.  Gra- 
ham Smith.  The  results  are  given  below,  and  it  will  be  seen 
that  they  are  very  much  below  those  given  by  experimenters 
on  small  pieces. 


6/2 


APPLIED  MECHANICS. 


Kind  of  Timber. 

Breadth 
and 
Depth. 

Span. 

How 

Loaded. 

Breaking- 
Weight. 

Modulus 
of 
Rupture. 

in. 

ft. 

American  red  pine 

I2.O  X    I2.O 

15.00 

Centre 

33497 

5238 

"           "      " 

12.0   X    12.0 

15.00 

« 

29908 

4680 

«<           «      « 

6.0  X    6.0 

7-50 

« 

7370 

4608 

Memel  fir      ... 

13-5  X    13-5 

10.50 

Distributed 

68560 

5274 

«       « 

13-5  x   13-5 

10.50 

« 

68560 

5274 

Baltic  fir  .... 

6.O  X    I2.O 

12.25 

Centre 

I9HS 

4878 

"        "    .     .     .     . 

6.O  X    I2.O 

12.25 

« 

23625 

6020 

Pitch  pine     .     .     . 

6.0  X    12.0 

12.25 

u 

23030 

5868 

«        « 

6.0  X    12.0 

12.25 

" 

23700 

6048 

«         « 

14.0  x  15.0 

10.50 

tt 

134400 

8064 

«         « 

14.0  x  15.0 

10.50 

« 

132610 

7956 

Red  pine  .... 

6.O  X    I2.O 

12.25 

u 

16800 

4284 

«       « 

6.0   X    12.0 

12.25 

« 

19040 

4860 

Quebec  yellow  pine 

14.0  x  15.0 

10.50 

Distributed 

68600" 

4122 

«           «         « 

14.0  x  15.0 

10.50 

" 

68600 

4122 

<<           «         « 

14.0  x  15.0 

10.50 

Centre 

85792 

5H8 



14.0  x  15.0 

10.50 

« 

76160 

4572 

Two  tests  by  R.  Baker  are  also  mentioned  by  D.  K.  Clark. 

Bauschinger  also  made  quite  an  extensive  series  of  tests  of 
German  woods,  an  account  of  which  will  be  given  later  on. 

A  great  many  tests  of  the  strength  and  stiffness  of  full- 
size  beams  of  spruce,  yellow  pine,  oak,  and  white  pine,  both 
under  centre  loads  and  distributed  loads,  have  been  carried 
on  in  the  Laboratory  of  Applied  Mechanics  of  the  Massa- 
chusetts Institute  of  Technology.  Tests  have  also  been  made 
upon  the  effect  of  time  on  the  stiffness  of  such  beams,  also 
on  the  strength  of  built-up  beams,  and  of  floors  and  fram- 
ing-joints, all  full  size.  A  summary  of  the  results  obtained 
will  be  given,  and  conclusions  drawn  as  to  the  proper  values 
of  the  modulus  of  rupture  and  modulus  of  elasticity,  etc.,  to 
be  used  in  practice. 

Before  giving  this  summary,  the  following  explanation  and 


TRANSVERSE  STRENGTH  OF   TIMBER.  6/3 

formulae  will  be  appended  for  the  convenience  of  those  who 
may  not  have  read  the  former  part  of  this  book.  The  different 
tables  of  results  given  in  different  handbooks  differ  in  the  form 
of  the  constant.  Thus,  the  constant  given  by  Trautwine  and 
Hatfield  is  one-eighteenth  the  modulus  of  rupture,  or  the  hypo- 
thetical breaking  centre  load  of  a  beam  one  inch  square  and 
one  foot  long  supported  at  the  ends  ;  while  Rodman  gives  one- 
sixth  of  the  modulus  of  rupture,  or  the  hypothetical  breaking- 
load  at  the  end  of  a  cantilever  one  inch  square  and  one  inch 
long. 

The  formulae  for  breaking-load  in  terms  of  the  modulus  of 
rupture,  and  for  modulus  of  rupture  in  terms  of  the  breaking- 
load,  for  some  of  the  most  usually  occurring  cases  of  rectan- 
gular beams,  are  appended. 

Let  W  —  breaking-load  in  pounds. 

/  =  modulus  of  rupture  in  pounds  per  square  inch. 
b  •=.  breadth  of  beam  in  inches. 
h  =  depth  of  beam  in  inches. 
/  =  length  of  beam  in  inches. 

Then  we  shall  have  :  — 
(a)  Beam  fixed  at  one  end  and  free  at  the  other. 

i°.  Single  load  at  free  end, 

w=f**  t 

'  6/' 
2°.  Load  uniformly  distributed, 


(b)  Beam  supported  at  both  ends. 
i°.  Single  load  at  the  middle, 


bh* 


_ 


6/4  APPLIED  MECHANICS. 

2°.  Load  uniformly  distributed, 

Hr_V**  /-3»? 

-  »T»  -  4  **•' 

3°.  Single  load  at  a  distance  a  from  the  origin, 


DEFLECTION   OF   BEAMS. 

While  the  preceding  formulae  refer  to  the  breaking-strength 
of  beams,  it  is  better  engineering  to  determine,  as  the  safe  load 
of  a  timber  beam,  the  load  that  will  not  deflect  it  more  than  a 
certain  small  fraction  (^  or  ¥^)  of  the  span. 
Let  W  =  given  load  in  pounds. 
b  =  breadth  in  inches. 
h  =  depth  in  inches. 
/  =  length  in  inches. 
v  =  greatest  deflection  in  inches. 
E  =  modulus  of  elasticity  of  the  material  in  pounds 
per  square  inch. 

(a)  Beam  fixed  at  one  end  and  free  at  the  other. 
i°.  Single  load  at  free  end, 

= 
2°.  Load  uniformly  distributed, 


_  3  =  3       _ 

=  ~  ~  =  2  vbh* 


(b)  Beam  supported  at  both  ends. 
i°.  Single  load  at  the  middle, 


-  = 

'  4  Ebh?  4  vbhf 


TRANSVERSE   STRENGTH  OF   TIMBER.  6?$ 

2°.  Load  uniformly  distributed, 


.   5 
32  32 

The  above  formulae  enable  us  to  determine  the  deflection 
of  a  beam  under  a  given  load  when  the  modulus  of  elasticity 
of  the  material  is  known,  or  to  determine  the  modulus  of  elas- 
ticity of  the  material  from  the  observed  deflection. 


^  £          LONGITUDINAL  SHEARING. 

In  any  rectangular  beam,  the  greatest  intensity  of  the  lon- 
gitudinal shearing-force  at  any  section  (i.e.,  its  intensity  at  the 
neutral  axis  of  the  section)  is,  if  we  let 

F  —  shearing-force  at  the  section,  technically  so-called, 

in  pounds, 

b  =.  breadth  of  beam  in  inches, 
h  =  depth  of  beam  in  inches, 


In  the  case  of  a  beam  supported  at  the  ends,  and  loaded  at 
the  middle  with  a  single  load,  we  have,  for  all  sections  except 

W 
the  middle,  F=  —  ;  and  hence,  if  we  denote  by  i  the  greatest 

intensity  of  the  longitudinal  shearing-force  at  the  neutral  layer, 
we  shall  have 


and  this  is  the  intensity  of  the  shearing-force  at  all  points 
along  the  neutral  layer,  except  at  the  middle  section,  where  it 
is  zero. 

In  the  case  of  a  beam  supported  at  the  ends,  with  the  load 
uniformly  distributed,  the  greatest  intensity  is  that  at  the  sup- 
port, and  is  also  given  by  equation  (i),  but  decreases  gradually 
to  the  middle. 


6;6 


APPLIED  MECHANICS. 


SUMMARY  OF   THE   TESTS. 

The  tests  recorded  may  be  divided  into  six  classes:  — 


i°.  Spruce  beams. 

2°.  Yellow-pine  beams. 

3°.  Time  tests. 


4°.  Oak  beams. 

5°.  White-pine  beams. 

6°.  Framing-joints. 


i°.  Spruce  Beams.  — Before  giving  a  summary  of  the  tests- 
made  in  this  laboratory,  I  will  insert  some  of  the  moduli  of 
rupture  and  moduli  of  elasticity  given  by  different  authorities. 

Moduli  of  rupture  are  given  as  follows  :  — 


Maximum. 

Minimum. 

Mean. 

Hatfield     .    .    .   ^ 

12996 

7506 

9900 

Rankine     .... 

12300 

9900 

IIIOO 

Laslett  

9707 

7506 

9045 

Trautwine  .... 

- 

- 

8lOO 

Rodman     .... 

- 

- 

6l68 

Hatfield's,  Laslett's,  Trautwine's,  and  Rodman's  figures  are 
from  their  own  experiments.  Trautwine  advises,  for  practical 
use,  to  deduct  one-third  on  account  of  knots  and  defects,  hence 
to  use  5400.  The  tables  show  the  values  obtained  in  these 
tests,  and  I  will  add  a  recommendation  as  to  the  values  of 
modulus  of  rupture  and  modulus  of  elasticity  suitable  to  use  in 
practice. 

As  a  result  of  the  tests  thus  far  made  in  my  laboratory, 
it  seems  to  me  safe  to  say,  if  our  Boston  lumber-yards  are 
to  be  taken  as  a  fair  sample  of  the  lumber-yards  in  the  case 
of  spruce,  —  if  such  lumber  is  ordered  from  a  dealer  of  good 
repute,  no  selection  being  made  except  to  discard  that  which  is 
rotten  or  has  holes  in  it,  —  that  3000  Ibs.  per  square  inch  is  all 
that  could  with  any  safety  be  used  for  a  modulus  of  rupture, 
and  even  this  might  err  in  some  cases  in  being  too  large ; 
(2°)  that,  if  the  lumber  is  carefully  selected  at  any  one  lumber- 


TRANSVERSE   STRENGTH  OF   TIMBER. 


yard,  so  as  to  take  only  the  best  of  their  stock,  it  would  not  be 
safe  to  use  for  modulus  of  rupture  a  number  greater  than  4000  ; 
and  if  we  required  a  lot  of  spruce  which  should  have  a  modulus 
of  rupture  of  5000,  it  would  be  necessary  to  select  a  very  few 
pieces  from  each  lumber-yard  in  the  city.  With  a  factor  of 
safety  four,  we  should  have  for  greatest  allowable  outside  fibre 
stress  in  the  three  cases  respectively,  750,  1000,  and  1250. 

The  modulus  of  elasticity  (i.e.,  that  determined  from  the 
immediate  deflections)  was:  maximum,  1588548;  minimum, 
-897961;  mean,  1332500. 

As  will  be  explained  when  the  results  of  the  time  tests  are 
given,  if  by  means  of  the  ordinary  deflection  formulae  we  wish 
to  compute  the  deflection  which  a  spruce  beam  will  acquire 
under  a  given  load  after  it  has  been  applied  for  a  long  time, 
we  should  use  for  modulus  of  elasticity  in  the  formulae  not 
more  than  one-half  of  the  values  given  above,  or  about  666300. 


6;8 


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APPLIED   MECHANICS. 


Yellow-Pine  Beams.  —  The  moduli  of  rupture  in  common, 
use  are  given  as  follows  by  different  authorities ;  viz.,  — 


Maximum. 

Minimum. 

Mean. 

Hatfield     .... 

2II68 

9000 

15300 

Laslett  

14162 

10044 

12254 

(  Yellow  pine 

9000 

Trautwine  .... 

_ 

(  Pitch  pine 

9900 

Rodman     .... 

9876 

8796 

9293 

A  summary  of  the  figures  obtained  from  these  tests  will  be 
given  in  a  table  at  the  end  of  these  remarks. 
It  will  be  observed  that  we  have  for 


Maximum. 

Minimum. 

Mean. 

Modulus  of  rupture  .     . 

II360 

3963 

7486 

Modulus  of  elasticity     . 

2386096 

1162467 

1757900 

If  by  means  of  the  ordinary  deflection  formulae  we  wish  to 
compute  the  deflection  which  a  yellow-pine  beam  will  acquire 
under  a  given  load  after  it  has  been  applied  for  a  long  time, 
we  should  use  for  modulus  of  elasticity  in  the  formulae  about 
one-half  of  the  values  above,  or  about  878950  (see  report  of 
time  tests). 


TRANSVERSE  STRENGTH  OF  TIMBER. 


683 


YELLOW-PINE   BEAMS. 


No. 
of 
Test. 

Width  and 
Depth. 

Span. 

Manner  of  Loading. 

Breaking- 
Weight, 
in  Ibs. 

Modulus  of 
Rupture. 

Modulus  of 
Elasticity. 

inches. 

ft.      in. 

3° 

3      X  i3| 

14      o 

Load  at  centre 

I5I58 

6614 

1937025 

32 

4-rV  X  I2-& 

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7383 

1733976 

33 

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18      o 

<«            « 

9832 

5386 

1793923 

47 

3      x  13! 

14      o 

«<            « 

19574 

8696 

2386096 

50 

4      X  i4lV 

21         0 

««            « 

12875 

59H 

1256286 

53 

3i    x  14 

24      6 

<«            «< 

10076 

7206 

1784426 

54 

3      X  i2| 

24      o 

«<            « 

9576 

9380 

2II682I 

56 

3i    x  14 

!5      4 

«            « 

10572 

4764 

1490396 

57 

2|f  X   12 

19         2 

«            it 

8472 

6950 

1444521 

59 

9     x  i3i 

24      o 

«            «« 

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5352 

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62 

4i    x  i2i 

19     10 

«            « 

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9102 

2037939 

63 

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<«            « 

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1966717 

67 

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18      6 

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7277 

1787610 

68 

4        X   I2| 

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10872 

2381685 

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20        0 

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I455308 

79 

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T  Load  equally  ) 

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4i    X  i2i 

17       4 

distributed  > 

16025 

4349 

1162467 

(      at  1  2  points) 

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4      X  12 

19      8 

Load  at  centre 

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9671 

1607336 

84 

4i    x  i2i 

21      4 

<«            <« 

"374 

6985 

1501854 

85 

4      X  iif 

20      6 

«            « 

16874 

11360 

2246154 

684 


APPLIED  MECHANICS. 


YELLOW-PINE   BEAMS.— Concluded. 


No. 

of 
Test. 

Width  and 
Depth. 

Span. 

Manner  of  Loading. 

Breaking- 
Weight? 
in  Ibs. 

Modulus  of 
Rupture. 

Modulus  of 
Elasticity. 

inches. 

ft.   in. 

87 

4   X  12* 

21   4 

Load  at  centre 

II272 

7335 

1535647 

88 

6   X  12* 

20   4 

tt      t 

15283 

6112 

1613012 

91 

4   X  12 

19  10 

«      t 

18074 

11303 

2223795 

92 

6   X  12 

6   5 

tt      t 

38090 

5092 

144 

4*  X  12 

21    0 

<t      t 

9433 

6012 

1628100 

145 

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6400 

1472000 

146 

4   X  12 

19   4 

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8060 

1839700 

147 

4   X  12 

19    2 

tt      t 

16748 

10032 

2286000 

216 

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15   8 

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15453 

7040 

1557300 

218 

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15   6 

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7484 

2OIO3OO 

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15   6 

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16632 

7427 

1623800 

220 

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15   6 

<t      tt 

I77IO 

7982 

I775IOO 

221 

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16   o 

t      tt 

16330 

7836 

I93I200 

222 

4$  X  12* 

16   o 

(           tt 

I85I5 

8884 

1786000 

223 

4   X  12 

17   o 

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13492 

7167 

1638500 

224 

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17   o 

t      tt 

16426 

8958 

1938900 

225 

4$  X  12 

15   6 

t       tt 

18705 

8786 

1729900 

226 

4$  X  12 

15   6 

f        ft 

16594 

7794 

I6II2OO 

252 

3H  X  iitf 

17   o 

t        It 

II006 

6002 

I27IOOO 

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4   X  12 

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15226 

7613 

1617700 

255 
257 

4&  X  I2f 

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15   6 
16   o 

t         t 

19425 
17424 

7975 
8031 

2214400 

1789800 

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4T7*  X  12 

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It         t 

18319 

8256 

1830000 

26l 

4T*  X  12 

15   o 

tt         t 

I78l8 

7978 

1989700 

Total 

mean  of  all  test 

s  excludin 

g  Time  tests  (51  bea 

ms)  = 

7486 

1757900 

In  regard  to  the  modulus  of  rupture  to  be  used  in  practice 
for  yellow  pine,  I  should  say,  that,  for  the  modulus  of  rupture 
of  yellow  pine  of  fair  quality,  I  should  not  feel  justified  in  using 
a  number  greater  than  5000  Ibs.  per  square  inch,  especially  for 
large  sizes,  such  as  9  X  14  inches,  12  X  16  inches,  etc.  My 
reason  for  this  conclusion  is,  that,  although  the  average  modulus 
of  rupture  derived  from  the  tests  already  enumerated  is  7486, 
nevertheless  we  have,  in  the  case  of  beam  No.  59,  a  modulus 
of  rupture  of  5300,  notwithstanding  the  fact  that  this  beam 
was  quite  free  from  knots,  cracks,  crooked  grain,  and  other 
defects,  and  had  been  selected  by  a  builder  as  one  of  excep- 
tionally good  quality.  With  a  factor  of  safety  four,  we  should 
have  about  1200  as  our  greatest  allowable  outside  fibre  stress. 


TRANSVERSE  STRENGTH  OF   TIMBER. 


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TRANSVERSE  STRENGTH  OF  TIMBER. 


687 


WHITE-PINE  BEAMS. 


No.  of 
Test. 

Width  and 
Depth. 

Span. 

Breaking 
Centre 
Load,  in 
IDS. 

Modulus  of 
Rupture. 

Modulus  of 
Elasticity. 

Remarks. 

inches. 

ft.  in. 

94 

3   X  III 

15  8 

5088 

3613 

924252 

(  Pattern  stock. 

95 

3   X  13 

14  o 

12588 

7251 

1280832 

•<  Clear  piece. 

(  Seasoned  3  yrs. 

96 

3   X  13 

16  6 

9088 

5324 

1072889 

97 

3   X  II 

15  8 

6088 

4729 

978256 

98 

2f  X  9i 

16  o 

6088 

6415 

1234880 

99 

2|  X  13 

15  6 

5988 

3438 

1020390 

IOO 

3   X  9i 

16  o 

4288 

4330 

H65937 

102 

3   X  iof 

15  6 

4790 

3855 

999190 

103 

3   X  ii 

16  6 

6588 

5390 

1242649 

104 

3   X  ni 

15  6 

5088 

3739 

931760 

128 

6   X  12 

19  10 

12922 

5340 

1380660 

129 

6  X  12 

19  08 

15060 

6170 

1565000 

130 

6  X  i2i 

19  08 

12340 

4954 

I222IOO 

131 

6  X  12 

19  10 

13023 

538o 

1307900 

132 

6   X  12 

19  10 

6231 

2575 

II03500 

133 

6  X  12 

19  08 

12912 

5290 

I297OOO 

134 

6   X  12 

20  oo 

11254 

4689 

1345700 

137 

6TV  X  12^ 

19  08 

13650 

5478 

1367700 

138 

6  X  12 

19  09 

14010 

5765 

I247IOO 

140 

6   X  12 

19  09 

9761 

4016 

II05600 

244 

4A  X  I2ff 

16  oo 

10179 

4300 

948600 

245 

4i  X  I2| 

15  6 

12984 

5620 

I27IOOO 

247 

4T*  X  ioi 

15  8 

6770 

3638 

IIII900 

248 

4i  X  I2i 

15  8 

7790 

3549 

1057300 

279 

4  X  I2| 

15  6 

9085 

4222 

IO84OOO 

280 

4  X  i2i 

15  6 

7575 

3521 

854300 

28l 

3*  X  i2T<V 

15   0 

12070 

5547 

I28IOOO 

282 

3tt  X  12^ 

15   0 

8660 

3998 

IO53OOO 

• 

283 

3||  X  i2i 

15  o 

7182 

3285 

970300 

284 

4   X  I2| 

15  o 

5685 

2456 

873300 

285 

3tt  x  12^ 

15  6 

6385 

3031 

901500 

286 

3l  X  12& 

15  6 

11791 

5449 

1258700 

287 

4   X  I2& 

15  6 

8265 

3802 

1049500 

288 

3l  X  I2& 

15  6 

JI272 

5353 

1295382 

289 

3|  X  12^ 

15  6 

5671 

2806 

8253OO 

2g6 

4   X  I2& 

15  6 

5571 

2563 

727200 

315 

3i  X  12 

16  o 

7165 

3820 

H584II 

I 

Mean  of  all  tests 
tute,  except 

made 
time  t 

at  Insti-  | 
ests,    1 

4451 

I  I  22000 

It  would  seem  to  the  writer  that  about  the  same  modulus 
of  rupture  should  be  used  for  white  pine  as  for  spruce. 


688 


APPLIED  MECHANICS. 


KILN-DRIED  WESTERN  WHITE  PINE. 


Modulus  of 

Modulus  of 

No.  of 
Test. 

Depth  and 
Width. 

Span. 

Manner 
of 
Loading. 

Breaking 
Load  (Ibs). 

Rupture 
(Ibs.  per 
Sq.  In.). 

Elasticity 
(Ibs.  per 
Sq.  In.). 

Remarks. 

inches. 

ft.    in. 

206 

2TV  X  13! 

15  05 

Centre 

9325 

7014 

1505000 

207 

2        X   I2f 

13  oo 

8814 

6432 

II93000 

208 

2TV  X  12^ 

15  oo 

3420 

2836 

752300 

2IO 

2TV  X  I  If 

15  oo 

4518 

4284 

1241400 

212 

2TV  X  13 

15  06 

6120 

4898 

1099300 

213 

2TV  x  isH 

15  06 

.... 

.... 

1276300 

214 

2        X  I2TV 

15  06 

7420 

6430 

1166000 

237 

2&  X  12H 

15  04 

8225 

6478 

I23IOOO 

Mean  = 

5482 

1183037 

HEMLOCK. 


No.  of  Test. 

Depth  and 
Width. 

Span. 

Manner 
of 
Loading. 

Breaking 
Load  (Ibs.). 

Modulus 
of 
Rupture 
(Ibs.  per 
Sq.  In.). 

Modulus  of 
Elasticity 
(Ibs.  per 
Sq.  In.). 

Remarks. 

inches. 

ft.   in. 

154 

155 

3i    X"f 
3TV  X  10 
2|    X    9| 

15  oo 

14  08 
12  09 

Centre 

6449 
4648 
4425 

3965 
4007 
3716 

870960 
97I7IO 

750400 

Nos.  154-160  are 
Eastern  hemlock 
seasoned  about  i 
year.  The  re- 

157 

II    10 

3223 

2381 

770900 

mainder  of  the 

158 

sS  X  IOA 

3      X    9f 

14  oo 
13  06 

4137 
2939 

3151 
2570 

735800 
833000 

lemlock  tests  are 
from  hemlock  cut 
in  Vermont  June, 

1  60 

2|     X  Hi 

12    08 

9433 

59" 

1086600 

1886;  first  growth 

177 
178 
1  70 

4*    X  12 
4     X  12 
4*    X  iiH 

15    08 

17  oo 

15    08 

9605 
5502 
3192 

4560 
2923 
1531 

1081100 
821990 
688960 

grown  on  high 
ground.  Sawed 
Sept.  25,  1886.  Re- 
ceived at  Institute 

M  M  M 
OO  CO  CO  • 
M  M  O  v 

4      X  i2i 
4      X  nil 
4      X  12 

15    06 
15    08 
15    06 

4584 
5133 
11073 

2175 
2539 
5363 

926560 

758390 
1296600 

Nov.  15,  1886.  Test- 
ed Dec.  2,  1886,  to 
March  9,  1887. 

I83 

4      X  12 

15    06 

13274 

6499 

1269800 

I84 

3l    X  iif 

15    08 

9964 

5142 

'1075600 

3|    X  nil 

17    02 

3679 

2059 

412670 

1  86 

15    08 

•     ' 

12488 

6535 

1327200 

Mean  = 

3825 

922250 

H4 

7f    X  io| 

20  04 

Centre 

19244 

8243 

2011188 

Yellow  birch.N.H. 

120 

7i   X  ioi 

19   06 

tt 

16150 

7627 

1583201 

ii           ii        it 

3VX     -Tf) 

13   06 

« 

14.065 

I2I22 

I  N.  H.  ash,  sea- 

127 

X  10 

)  soned  2  yrs. 

TRANSVERSE   STRENGTH  OF   TIMBER. 


689 


TIME    TESTS. 

The  following  is  a  record  of  the  time  tests  made  at  the 
Institute,  and  at  the  close  will  be  found  a  statement  in  regard 
to  the  proper  value  of  modulus  of  elasticity  for  use  in  com- 
puting deflections. 

TIME    TEST    NO.     I. 

Spruce  from  Maine,  received  at  Institute  October  30,  1885. 
All  the  beams  when  received  were  green  lumber,  except  F, 
which  was  seasoned  on  the  wharf  about  six  months.  Beams 
A,  B,  C,  D,  E,  and  F  were  seasoned  under  a  centre  load  in 
the  laboratory  in  steam  heat  from  November  10,  1885,  to  May 
8,  1886.  Beams  G,  H,  I  were  seasoned  in  the  same  room, 
without  load ;  span  =  20  feet  for  all  the  beams  under  load. 


Beam. 

A  (164) 

B  (163) 

C  (162) 

D  (161) 

E  (169) 

F  (168) 

Description  of  lumber,  . 
Dimensions  •<  2!,1^ina  '  ' 

clear 
4"  x  12" 
3f'xnf" 

knotty 
4"  x  12" 
4"  x  i  if" 

knotty 
4"  x  12" 
Sf'xiii" 

clear 
4"  X  I2//// 

clear 
6"  xi  a" 
5f"  x  iif" 

clear 
6//  X  ""// 

f  with  wt.  of 

Max.  fibre  1      beam,     . 
stress,  Ibs.  -|  without 

1078 

1076 

1074 

1070 

"33 

1136 

per  sq.  in.         wt.     of 

I.     beam,     . 
Deflection,  load  first  ap- 

1003 

1003 

1003 

o"  .  7675 

1003 
o".6io8 

1057 
o".7i88 

1051 
o".5454 

Deflection    at    end    of 

test,  
E  (immediate),      .    .    . 

1462200 

i  ".4707 
1262900 

i".5I54 
1045700 

1-3734 
1314000 

1.2667 
1175000 

0.3151 
1540000 

E  (final) 

789000 

546000 

52QOOO 

584000 

667000 

2666000 

6574 

7260 

5007 

0°4^AJW 
6066 

6779 

8574 

lbsPtUrper  J  v^tthoutof 

sq.  in.          ^    beam, 

6500 

7187 

4937 

6OOO 

6708 

8500 

Weight    per  cu.   ft.  at 

beginning,  Ibs.,      .     . 

35-5 

34-9 

33-6 

3L9 

34-i 

35-4 

Weight   per  cu.   ft.  at 

end   Ibs  ,   

27.  I 

28.1 

2  e  ,  e 

25  .  7 

28.2 

29  6 

Date  of  testing,     .    .     . 
E  (after  seasoning),  Ibs. 

May  ii,  '86 

May  ii,  '86 

May  io,  '86 

May  io,  '86 

May  14,  '86 

May  13,  '86 

per    sq.    in.    of    final 

section 

1866900 

1509500 

1367300 

1625500 

1737400 

1718300 

Average  E  (immediate)  =  1300000. 
Average  E  (final)  =    963500. 

Average  modulus  of  rupture  beams  under  load  =  6710. 
All  quantities  in  the  above  table  except  the  last  were 
section. 


calculated  by  using  the  original 


690 


APPLIED  MECHANICS. 


Beam. 

G  (165) 

H  (166) 

I  (167) 

clear 

clear 

knotty 

Dimensions  -1  - 

3J"  X  II$" 

3f  "  x  12" 

Modulus  of  rupture,  <  with  wt.  of  beam,   .    .    . 
Ibs.  per  sq.  in.            \  without  wt.  of  beam,  .     . 
Weight  per  cu.  ft.  at  beginning,  Ibs.,     .... 
Weight  per  cu.  ft.  at  end,  Ibs.,    
Date  of  testing  

6525 

27-3 
May  ii,  '86 

7187 

27 

May  12,  '86 

3oI2 
27.2 

May  12,  '86 

E  (after  seasoning),  Ibs.  per  sq.  in.  of  final  section, 

1603700 

1748900 

1457000 

Average  E  (final  section),  beams  without  load, 1603200 

Average  modulus  of  rupture  (original  section) 6508 


TIME   TEST   FOR  SHORT   PERIODS   OF  TIME. 


Total  in- 

No. 

Depth  and 
Width. 

Span. 

Original 
Modulus  of 
Elasticity. 

creased 
Deflec- 
tion. 

Modulus 
of 
Rupture. 

Description  of  Test. 

inches. 

ft.  in. 

inches. 

Ibs. 

108 

6X  12 

17    04 

1269670 

.1241 

5066 

Spruce-beam  load  equally  dis- 

tributed at  12  points.    The 

beam  was    subjected    to  a 

load  of  5031  Ibs.  for  898  hrs. 

148 

4i35Xi4| 

17    04 

1211800 

•3765 

4668 

Yellow-pine  beam,   load  dis- 

tributed   equally    over    12 

points.     The  beam  was  sub- 

jected to  a  load  of  6355  Ibs. 

for  29  days. 

TIME    TEST    NO.    2. 

Spruce  beams  cut  in  Maine  in  the  spring  of  1886.  Received 
at  Institute  September  13,  1886.  Beams  A,  B,  C,  D,  E,  and 
F  were  seasoned  under  a  centre  load,  in  the  laboratory,  in 
steam  heat,  from  September  15,  1886,  to  April  2,  1887  (2O° 
days).  Beams  G,  H,  I  were  seasoned  in  the  same  room,  but 
were  not  loaded.  All  the  beams  were  without  load  between 


TRANSVERSE   STRENGTH  OF   TIMBER. 


691 


April  2  and  date  of  testing.     Span  =  1 8'  oo"  for  all  beams 
under  load. 


Beam. 

A  (194) 

B  (193) 

C  (191) 

D  (190) 

E  (195) 

F  (192) 

Description  of  lum- 
ber,      

clear 

clear    -j 

cross- 
grained 

j-  knotty  j 

straight- 
grained, 
some  knots 

straight- 
grained, 
some  knots 

Dimen-  j  original,  . 
sions  1  final,    .     . 

sirxulV 

3*"xn&" 

4*"   xxxjf" 

5*"xxxf" 

5H"x«x«" 

f  with 

wt.  of 

Max.  fibre 

beam, 

1020 

IOII 

1136 

994 

1095 

1093 

stress,  Ibs. 

•{  with- 

per sq.  in. 

out 

wt.  of 

L  beam, 

966 

956 

1075 

943 

1037 

1041 

Deflection,load  first 

applied, 

.     .    .     . 

o"-37°7 

o".42S3 

o"-5335 

°".4794 

o".5024 

o".5747 

Deflection  at  end  of 

• 

test 

o".9i8i 

o"  •  8447 

//    ___g 

l"   OA.Z1 

l"   OI7O 

J//      CQCQ 

E  (immediate),  .    . 

1689000 

1449000 

1333000 

I288OOO 

1327000 

II73000 

E  (final), 

682000 

730000 

577000 

591000 

655000 

448000 

Modulus 

with 
wt.  of 

of  rup-_ 
turc  Ibs   - 

10283 

7838 

3243 

5038 

7448 

5Il6 

without 

per  sq. 

wt.  of 

in. 

beam, 

10229 

7783 

3182 

4987 

7392 

5064 

Weight  per  cu.  ft. 

at  beginning,  Ibs., 

31-8 

32.8 

35-4 

30.0 

34-1 

30-5 

Weight  per  cu.  ft. 

at  end,  Ibs.,.    .    . 

28.6 

26.1 

29.1 

27.0 

27.7 

26.3 

Date  of  testing, 

Apr.  25,  '87 

Apr.  19,  '87 

Apr.  5,  '87 

Apr.  4,  '87 

Apr.  27,  '87 

Apr.  8,  '87 

E  (after  seasoning), 

Ibs.  per  sq.  in.  of 

final  section,    .    . 

2125500 

1852300 

1731600 

1517800 

1662100 

1294100 

Average  E  (immediate)  =  1376500. 
Average  E  (final)  =    614000. 

Average  modulus  of  rupture  beams  under  load 


6494. 


692 


APPLIED   MECHANICS. 


Beam. 

G  (188) 

H  (187) 

I  (189) 

knotty 

clear 

knotty 

Dimensions  -|final  

3f"xn|" 

3j|"  x  n£" 

4"XI2" 

3f"xn{|" 

Modulus  of  rupture,  j  with  wt.  of  beam,  .    .    . 
Ibs.  per  sq.  in.           1  without  wt.  of  beam,  .    . 
Weight  per  cu.  ft.  at  beginning,  Ibs.,  

5218 
5167 
29.5 

8667 
8598 
35.8 

4796 
4751 

Wt  per  cu  ft  at  end,  Ibs.,     

24.9 

29.3 

24  8 

Date  of  testing  .'    

Mch.  21,  '87 

Apr.  i,  '87 

Mch.  16,  '87 

E  (after  seasoning),  Ibs.  per  sq.  in.  of  final  section, 

*3553°° 

1914500 

1573600 

Average  E  (final  section),  beams  without  load  =  1614500.      t 
Average  modulus  of  rupture  (original  section)  =       6227. 

TIME    TEST    NO.    3. 

Yellow-pine  beams  from  Georgia,  cut  in  season  of  1886. 
Received  at  Institute  September  13,  1887.  The  lumber  was 
purchased  in  sticks  of  double  length,  and  cut  in  two  for  testing. 
The  numbers  indicate  the  stick,  and  the  letter  "  B  "  the  butt, 
and  "T  "  the  top  end  of  the  same.  Beams  I  T,  2  B,  3  B,  4  T, 
5  B,  5  T  were  seasoned  under  load,  the  remainder  being  sea- 
soned without  load.  Span  =  i%. 


TRANSVERSE  STRENGTH  OF  TIMBER. 


693 


Beam. 

iT(238) 

2  B  (239) 

3  B  (240) 

4  T  (241) 

5  B  (242) 

5  T  (243) 

Description    of  j 
lumber,     .    .  | 

clear, 
cross- 
grained 

clear, 
some  dry 
rot 

clear, 
straight- 
grained 

clear, 
cross- 
grained 

clear, 
straight- 
grained 

clear, 
straight- 
grained 

Dimen-  {  original 

4iV  x  i23V' 

43£"  X  12/g" 

4V  x  i23y 

3tt"xi2&" 

6&"xiiff" 

65%"  X  12$" 

sions  1  final, 

4"     x  nil" 

4ffe"  x  I2iV  ' 

3tt"xnH" 

3tt"xnH" 

6"xxx&" 

6£"  x  iilj" 

f  with 

Max. 

wt.  ol 

fibre 

beam, 

1358 

1259 

1381 

1435 

1626 

1509 

stress,    - 

with- 

Ibs. per 

out 

sq.  in. 

wt.  of 

„  beam, 

1283 

"93 

1302 

1364 

1S3I 

1426 

Deflection,     load 

first  applied,     . 

o".53io 

o".55i8 

o//.47oo 

o".s6i8 

o"-443i 

o".4836 

Deflection  at  end 

of  test, 

.    .    . 

o".98i7 

l".2387 

o".9424 

i".4566 

o".8364 

o".8449 

E  (immediate),    . 

1536000 

1351000 

1763000 

1545000 

2290000 

1889000 

E  (final), 

.    .    . 

832000 

602000] 

879000 

596000 

1213000 

1081000 

Modu- 

'  with 

lus  of 

wt.  of 

rup- 

beam, 

6290 

6234 

9545 

6570 

10630 

1  0000 

with- 

ture, 

Ibs.  per 

out 
wt.  of 

sq.  in. 

beam, 

6213 

6163 

9461 

6495 

10540 

9924 

Weight  per  cu.  ft. 

at     beginning, 

Ibs, 

45.6 

40.6 

48.3 

._  x 

,,-  8 

Weight  per  cu.  ft. 

' 

45  *° 

at  end, 

Ibs.  .    . 

38.4 

36.8 

43-3 

39-o 

45-8 

41.4 

Date  of  testing,  . 

May  22,  '88 

May  28,  '88 

May  28,  '88 

May  28,  '88 

May  31,  '88 

June  i,  '88 

E  (after 

season- 

ing),    Ibs.    per 

sq.  in. 

of  final 

section, 

.    .    . 

1898000 

1651000 

2240000 

1803700 

2501400 

2103400 

I 

Average  E  (immediate)  =  1729000. 
Average  E  (final)  =    867000. 

Average  modulus  of  rupture  beams  under  load  =  8211. 


694 


APPLIED  MECHANICS. 


Beam. 

i  B  (229) 

2  T  (232) 

3T(233) 

4  B  (228) 

/       clear, 

clear, 

K     straieht- 

dry  rot 

unsound 

grained 

grained 

(  original,  
Dimensions-:  , 

4±"  X  I23y 

43V'  x  "t" 

43V'  X  12" 

f  with     wt.     of 
Modulus  of  rup-         , 

ture,    Ibs.    Per     withou;;t'0' 

941? 

3069 

5838 

9960 

Sq'  m"                  I     beam,   /   .    . 

9330 

3007 

5754 

9892 

Weight  per  cu.  ft.  at  beginning, 

Ibs.,  

46.1 

Weight  per  cu.  ft.  at  end,  Ibs., 

41.4 



March  28,  '88 

April  6,  '88 

April  9,  '88 

March  26,  '88 

E  (after  seasoning),  Ibs.  per  sq. 

in.  of  final  section,      

2141400 

1393000 

1951000 

1895600 

Average  E  (final  section),  beams  without  load  =  1820200. 
Average  modulus  of  rupture  (original  section)  =       7071. 


TIME    TEST    NO.    4. 

Spruce  from  Maine.  Green  lumber.  Received  at  Insti- 
tute September  n,  1888.  Beams  A,  B,  C,  D,  J,  K  were  sea- 
soned under  load  in  the  laboratory  in  steam  heat  from  Sep- 
tember 13,  1888,  to  March  25,  1889  (194  days).  Beams  E,  F, 
G,  and  H  were  seasoned  in  the  same  room,  but  not  loaded, 
All  the  beams  were  without  load  from  March  25,  1889,  to  tne 
date  of  testing.  Span  =  18'  for  the  beams  while  under  load. 


TRANSVERSE   STRENGTH  OF   TIMBER. 


695 


Beam. 

A  (273) 

B  (274) 

C  (276) 

D  (278) 

J  (277) 

K  (275) 

Description  of  lum- 

{knotty, 

clear, 

clear, 

ber,  .    . 

ave.  stock 

ave.  stock 

straight- 

straight- 

straight- 

grained 

grained 

grained 

Dimen-  j  original,     . 

4^"x  nj|" 

4iV'x  IITB" 

4A"xnf" 

4i"xn|" 

5ii"x  ii$|" 

5*"XT2" 

sions  1  final,     .     . 

4"x  ni" 

3t|"xiiTy 

4"  XI  if" 

31ITXII&" 

SrV'xnf" 

5f"xnTy' 

•  with 

Max.  fibre 

wt.   of 

stress, 

beam, 

1422 

1442 

1444 

1436 

1750 

1705 

Ibs.    per  " 

without 

sq.  in. 

wt.  of 

beam, 

1364 

1385 

1387 

1379 

1698 

1645 

Deflection, 

load  first 

applied, 

.    .    .    . 

o".6699 

o".77i6 

o".66i8 

o"-7379 

o".8446 

o"-7439 

Deflection  at  end  of 

fdCf 

"           <21 

i"  837? 

i"  ^•560 

//     fQ0^ 

I"-7359 

l"     A9&A 

E  (immediate),     .    . 

1326000 

*      •°j/O 

1169000 

1387000 

1224000 

1308000^ 

i       .4^04 

1431000 

E  (final), 

.    .    .    . 

682000 

491000 

597000 

532000 

637000 

745000 

Modulus 

with 
wt.  of 

of     rup- 
ture, Ibs.  - 

beam, 
without 

7545 

7211 

5597 

6622 

6484 

7099 

per     sq. 

wt.   of 

in. 

.   beam, 

7487 

7155 

5540 

6565 

6431 

7038 

Weight  per  cu.  ft.  at 

beginning,  Ibs.,    . 

34-3 

33-5 

33-o 

33-6 

3T-5 

36-4 

Weight  per  cu.  ft,  at 

end,  Ibs., 

.    .    .    . 

27-3 

25-4 

26.8 

26.9 

27.1 

28.6 

Date  of  testing,  .    . 

Mch.  26,  '89 

Apr.  3,  '89 

Apr.  9,  '89 

Apr.  n,.  '89 

Apr.  10,  '89 

Apr.  8,  '89 

E  (after  seasoning), 

Ibs.  per  sq.  in.  of 

final  section,     .    . 

1631700 

1405500 

1747900 

1503200 

1576000 

1722600 

Average  E  (immediate)  =  1307500. 
Average  E  (final),  =    614000. 

Average  modulus  of  rupture  beams  under  load  =  6760. 


696 


A  P  PLIED    ME  CIIA  NICS. 


Beam. 

E  (270) 

F  (271) 

G(272) 

H  (251) 

nearly 
clear, 
straight- 
grained 
3l"  x  12" 

4"xiiH" 
311"  x  n|" 
7615 

7552 
37-2 
27.0 
Mch.  19,  '89 

1771000 

3J$"x  12" 
85581 

8496 
36.8 
26.3 
Mch.  22,  '89 

1770000 

knotty 
43V'xii^' 
4732 

4676 
33-o 

Nov.  6,  '88 
1266500 

Dimensions  j  j>rigmal  

Weight  per  cu.  ft.  at  beginning,  Ibs.,    .    . 
Weight  per  cu.  ft.  at  end,  Ibs.,      .... 

3f"xii$" 
8293 

8237 
33-5 
27-3 
Mch.  13,  '89 

1632500 

Date  of  testing 

E  (after  seasoning),  Ibs.  per  sq.  in.  of  final 

Average  E  (final  section),  beams  without  load  =  1610000. 
Average  modulus  of  rupture  (original  section)  =        7300. 

DEFLECTIONS    WITH    TIME. 

From  the  above  it  is  plain  that  the  deflection  of  a  timber 
beam  under  a  long-continued  application  of  the  load  may  be 
2  or  more  times  that  assumed  when  the  load  was  first  applied ; 
and  in  order  to  compute  it  by  means  of  the  ordinary  deflection 
formulae,  we  should  use  for  E  not  more  than  £  the  value  de- 
rived from  quick  tests. 

LONGITUDINAL    SHEARING. 

Below  are  given  tables  showing  the  greatest  intensity  of 
the  shear  at  the  neutral  axis  of  each  beam  at  fracture  as  calcu- 
lated from  the  formula  on  page  675. 

TABLE  OF  BEAMS  WHICH   GAVE  WAY   BY    LONGITUDINAL  SHEARING. 


Spruce. 

Yellow  Pine. 

Oak. 

White  Pine. 

No. 

Intensity  of 
Shear,  Lbs. 
per  Sq.  In. 

No. 

Intensity  of 
Shear,  Lbs. 
per  Sq.  in. 

No. 

Intensity  of 
Shear,  Lbs. 
per  Sq.  In. 

No. 

Intensity  of 
Shear,  Lbs. 
per  Sq.  In. 

22 
24 
31 

46 
90 

32I 

202 
I90 
154 

3 

233 
273 
I30 

233 

3° 
32 
33 
50 
147 

273 
242 

177 
264 

109 
269 

152 
379 

129 

233 

119 
1  80 

Average  198. 

Average  222. 

Average  266. 

Average  151. 

TRANSVERSE   STRENGTH  OF   TIMBER. 


697 


TABLE  OF   BEAMS  WHICH   DID   NOT  FAIL   BY  SHEARING. 


Spruce. 

Yellow  Pine. 

Oak. 

White  Pine. 

Hemlock. 

Intensity  of 
No  1  Shear.  Lbs. 

No. 

Intensity  of 
Shear,  Lbs. 

No 

Intensity  of 
Shear,  Lbs. 

No. 

Intensity  of 
Shear,  Lbs. 

No 

Intensity  of 
Shear,  Lbs. 

per  Sq.  In. 

per  Sq.  In. 

per  Sq.  In. 

per  Sq-.  In. 

per  Sq.  In. 

3 

181 

47 

359 

48 

149 

94 

112 

154 

129 

4 

301 

53 

179 

51 

238 

95 

282 

"5 

5 

174 

54 

203 

55 

205 

96 

176 

156 

119 

6 

230 

56 

185 

80 

211 

97 

139 

86 

7 

308 

57 

182 

no 

182 

98 

164 

158 

97 

8 

170 

59 

133 

112 

137 

99 

121 

78 

9 

141 

62 

237 

113 

100 

III 

160 

215 

10 

106 

63 

211 

Iiq 

123 

102 

112 

177 

147 

ii 

208 

64 

161 

116 

203 

103 

I5I 

178 

88 

12 

126 
I05 

65 
67 

'83 
196 

«7 

118 

& 

104 
III 

'3 

179 
180 

5° 
72 

15 

304 

68 

278 

119 

193 

128 

136 

181 

82 

16 

277 

69 

"5 

122 

93 

I30 

129 

182 

174 

17 

465 

71 

227 

123 

158 

'31 

137 

^83 

209 

18 

202 

74 

175 

125 

161 

132 

66 

184 

163 

19 

138 

75 

233 

I49 

MS 

133 

136 

185 

61 

20 

160 

76 

233 

151 

137 

142 

186 

209 

21 
23 

III 

77 
78 

109 

;g 

III 

138 
245 

190 

25 

123 

81 

238 

i69 

150 

247 

106 

26 

82 

246 

171 

150 

248 

117 

27 

167 

84 

167 

172 

125 

279 

140 

28 

134 

271 

173 

280 

117 

29 

129 

87 

175 

174 

197 

281 

191 

37 

169 

88 

161 

178 

282 

135 

45 

1-33 

91 

282 

I77l 

245 

283 

112 

49 

205 

92 

397 

259 

204 

284 

87 

60 

406 

144 

260 

184 

285 

IOO 

66 

230 

MS 

191 

262 

184 

286 

182 

7° 

206 

146 

206 

264 

216 

287 

127 

72 

177 

218 

244 

265 

324 

288 

178 

196 

219 

244 

289 

9* 

197 

120 

220 

265 

.... 

296 

86 

198 

170 

221 

248 

199 

222 

285 

200 

IOI 

223 

212 

201 

191 

224 

264 

202 

160 

225 

285 

203 

189 

226 

253 

204 

1  88 

2S2 

177 

205 

128 

254 

240 

297 

97 

255 

267 

299 

232 

257 

256 

300 

298 

258 

260 

301 

295 

26l 

268 

305 

152 

307 

156 

308 

179 

309 

141 

311 

129 

314 

1  02 

3*5 

121 

316 

I84 

3*7 

175 

3'8 

191 

ojr 

201 

320 

85 

322 

MS 

Average  182. 

Average  231. 

Average  176. 

Average  134. 

Average  123. 

698  APPLIED  MECHANICS. 

One  would  naturally  expect  to  find  the  intensity  of  the 
shearing-stress  at  fracture  less  in  the  case  of  the  beams  that 
did  not  fail  by  shearing  than  in  the  case  of  those  that  did  ;  and 
this  is  seen  to  be'  generally  true  (making  allowance  for  different 
qualities)  both  in  the  case  of  spruce  and  hard  pine. 

The  notable  exceptions  seem  to  be,  in  the  case  of  spruce, 
beams  Nos.  4,  7,  15,  16,  17,  60,  all  of  which  have  this  intensity 
very  large.  If  these  be  omitted  from  the  list,  the  average  for 
those  that  did  not  give  way  by  shearing  would  be  163  pounds 
per  square  inch. 

In  the  case  of  yellow  pine,  the  notable  exceptions  are  beams 
Nos.  47,  77,  and  92 ;  and,  if  these  be  omitted,  the  average  for 
those  yellow-pine  beams  that  did  not  fail  by  shearing  would  be 
219  pounds. 

Moreover,  it  is  to  be  observed,  that,  in  the  case  of  the  spruce, 
Nos.  4,  7,  15,  1 6,  and  17  were  all  of  smaller  dimensions  than 
those  used  in  practice. 

For  breaking  shearing-strength  per  square  inch,  in  the  case 
of  each  wood,  it  seems  to  the  author  that  it  would  be  proper 
to  use  a  value  somewhere  near  the  lowest  of  those  given  in 
the  table  of  beams  which  gave  way  by  longitudinal  shearing. 

It  will  also  be  observed  that  these  shearing-forces  are  less 
than  those  obtained  from  the  experiments  on  direct  shearing 
along  the  grain,  made  at  the  Watertown  Arsenal ;  and  this  is 
naturally  to  be  expected,  for  the  shearing  in  their  case  took 
place  along  a  section  that  was  perfectly  sound,  while  in  these 
cases  it  took  place  at  the  weakest  point. 

6°.  Framing- Joints.  —  Another  matter  intimately  connected 
with  the  strength  of  timber  beams  is  the  strength  of  the  beam 
after  it  has  been  cut  in  some  of  the  various  ways  commonly 
employed  in  framing.  We  are  often  told  that  a  notch  cut  on 
top  of  a  beam,  or  at  the  middle  of  its  depth,  or  near  the  sup- 
port, does  but  little  injury ;  but  the  tests  made,  show  the  injury 
to  be  very  large,  amounting  to  a  reduction  of  the  strength  of 


TRANSVERSE  STRENGTH  OF   TIMBER. 


699 


FIG.  243. 


FIG.  244. 


the  beam  to  one-fourth  or  one-fifth  of  its  original  strength, 
with  some  of  the  most  approved  framing.  The  fact  is,  that, 
with  a  material  where  the  shearing-strength  along  the  grain  is 
so  small  as  it  is  in  the  case  of  timber,  almost  any  cutting  does 
a  great  deal  of  injury;  and  it  is  much  better  to  avoid  framing 
whenever  it  is  possible,  and  use  stirrup  irons  instead.  In  these 
tests,  only  two  of  the  most  approved  framing-joints  have  been 
tested ;  viz.,  the  joint  known  as  the  "  tusk-and-tenon,"  shown 
in  Fig.  243,  and  used  for  framing  the  tail-beams  of  a  floor  into 
the  headers,  and  the  "double 
tenon  and  joint  bolt,"  shown  in 
Fig.  244,  and  used  for  framing 
the  headers  into  the  trim- 
mers. 

The  arrangement  is  shown  in  plan  in  Fig.  245,  where  I  and 
2  are  the  trimmers,  3  is  the  header,  and  4,  5,  and  6  are  the 
tail-beams ;  the  latter  being  supported  at 
one  end  on  the  header,  and  at  the  other 
on  the  wall,  the  header  being  supported 
by  the  trimmers,  and  the  trimmers  being 
supported  on  the  walls  at  both  ends. 

It  is  sometimes  the  practice  to  hang 
the  header  in  stirrup  irons,  and  this  is  an 
improvement ;  but  it  is  very  seldom  that 
the  tail-beams  are  hung  in  stirrup  irons, 
and  these  tests  have  shown  the  weakening 
already  referred  to,  from  the  mortises  cut 
in  the  header  to  admit  the  tail-beams. 

A    spruce    floor    was    first    built    and 
tested,  the  following  being  a  partial  ac- 
FIG.  245.  count  of  the  test :  — 

No.  52.  —  Section  of  a  floor  between  the  trimmers.  Spruce  : 
three  tail-beams,  2  inches  by  12  inches  each,  framed  into  a  31- 
inch  by  iif-inch  header;  header  in  turn  framed  into  sections 


^                    ! 

-4U           / 
j!  i            ( 

i 

>                4 

i 

\               5 

i 

3 

S           6 

f 

il 

J                        2 

QL 

i 

7OO  APPLIED   MECHANICS. 

of  the  trimmers  by  double  tenon  and  joint-bolt,  cross-bridged  in 
two  places ;  tail-beams  framed  by  tusk-and-tenon  joint,  pinned, 
floored  over  and  furred  below ;  load  at  centre,  distributed  be- 
tween the  three  tail-beams  by  bridging. 

Span  =  1 6  feet ;  weight  of  joist,  flooring,  etc.,  =  331  Ibs. 

11238  Ibs.  =  breaking-load. 

Joist  on  east  side  broke  by  splitting  off  at  the  tenon,  bore 
7988  Ibs.  after.  The  load  was  then  increased.  Centre  tail-beam 
broke  by  tension  at  9988  Ibs.,  on  account  of  cross-grain  in  the 
lower  fibres.  A  split  also  started  at  the  lower  tenon  of  the 
header,  which  at  the  time  of  breaking  was  rapidly  increasing. 

Average  modulus  of  rupture  of  the  tail-beams,  including 
their  own  weight,  etc.,  —  3801  Ibs.  per  square  inch. 

Average  modulus  of  elasticity  of  tail-beams  =  1399141  Ibs. 
per  square  inch. 

It  is  to  be  noticed,  that  the  header  already  began  to  crack 
when  the  tail-beams  broke,  and  hence  that  the  floor  could  have 
borne  but  little  more,  even  if  the  load  had  been  uniformly  dis- 
tributed :  hence  that,  in  this  case,  the  breaking-strength  of  the 
floor  would  be  determined  by  calculating  the  loads  at  the  centre 
of  the  tail-beams,  instead  of  accounting  it  as  distributed  ;  in 
other  words,  the  breaking-weight  would  be  about  one-half  what 
we  should  get  by  considering  the  load  as  distributed  on  the 
tail-beams. 

YELLOW-PINE    HEADERS. 

Six  tests  of  yellow-pine  headers  have  been  made  at  the 
Massachusetts  Institute  of  Technology,  and  the  results  will  be 
given  here. 

It  will  be  seen  from  these  tests,  that  the  first  of  these  head- 
ers had  for  its  breaking-weight  10916  Ibs.,  and  the  second 
13163,  or  in  each  case  one-half  the  load  on  the  floor.  To 
institute  a  comparison,  we  may  observe  that,  if  a  6-inch  by 
12-inch  yellow-pine  header  6  feet  8  inches  long,  with  four  tail- 


TRANSVERSE   STRENGTH  OF   TIMBER. 


701 


beams  18  feet  long,  were  to  support  a  floor,  the  floor  surface 
would  be  96  square  feet,  giving  48  square  feet  to  be  supported 
by  the  header.  This,  if  the  floor  were  loaded  with  100  Ibs.  per 
square  foot,  would  bring  upon  the  header  4800  Ibs.,  or  about 
one-half  the  breaking-weight  of  a  header  only  5  feet  4  inches 
long  ;  whereas,  it  would  commonly  be  supposed,  that,  with  such 
a  construction  for  100  Ibs.  per  square  foot  of  floor,  we  should 
have  provided  an  unnecessarily  large  margin  of  safety. 

As  to  the  fact  that  the  header  supported  in  stirrup  irons 
bore  less  than  that  which  was  framed,  this  must  be  due  to  a 
difference  in  the  quality  of  the  timber ;  and  it  would  be  unfair 
to  conclude  from  only  two  tests  that  the  second  was  a  stronger 
mode  of  construction  than  the  first,  even  as  far  as  the  header 
itself  is  concerned. 

The  fact,  also,  that  a  6-inch  by  1 2-inch  yellow-pine  beam  5 
feet  4  inches  long  bore  48000  Ibs.  centre  load,  equivalent  to 
96000  distributed,  without  breaking,  while  the  header  broke  at 
10916,  shows  what  an  enormous  weakening  is  caused  by  cutting 
mortises,  and  how  much  strength  would  be  gained  by  avoiding 
all  framing,  and  using  stirrup  irons  to  support  the  tail-beams  in 
all  cases  where  they  cannot  be  supported  on  top  of  the  header 
bearing  the  latter. 

No.  83.  —  Test  of  yellow-pine  headers. 


FIG.  '246. 

The  headers,  6  inches  by  12  inches,  span  5  feet  4  inches, 
were  hung  at  either  end  in  iron  stirrups,  from  trimmers  6 
inches  by  12  inches  by  20  inches,  which  in  turn  were  sup- 


7O2  APPLIED  MECHANICS. 

ported  on  jack-screws.  The  headers  were  mortised  in  three 
places  (16  inches  on  centres)  for  three  3-inch  by  1 2-inch  yellow- 
pine  tail-beams  10  feet  in  length. 

The  load  was  applied  at  the  centre  of  the  tail-beams,  and 
divided  equally  among  the  three  by  iron  bridging.  v  ' 

The  tail-beams  were  cross-bridged  in  two  places,  6  feet 
apart,  by  2-inch  by  3-inch  spruce  bridging,  and  also  floored  over 
with  yellow-pine  flooring  I  inch  thick. 

The  weight  of  the  tail-beams,  bridging,  flooring,  etc.,  which 
was  supported  by  header,  was  833  Ibs. 

Weight  of  north  header,  106  Ibs. 

Weight  of  south  header,  122  Ibs. 

The  headers  were  heart  pieces,  coarse  grain,  and  sappy. 
The  north  one  had  a  few  season  cracks  extending  from  mortise 
to  mortise  on  the  outside.  The  tail-beams  were  of  medium 
quality,  two  of  them  having  sapwood  on  the  edges :  the  third 
was  much  coarser,  but  contained  more  pitch,  being  heavy. 
Weight  of  yoke  and  iron  bridging,  366  Ibs. 

Details  of  the  Test. 
15366  Ibs.     Cracks  heard  in  north  header. 

19866    "       Loud  cracks;  season  cracks  opening  in  north  header. 
21366    "       North  header  giving  way,  load  dropped  500  Ibs.;   cracks 

§  inch  wide. 

26366    "       South  header  began  to  show  cracks. 
26366    "       Centre  tail-beam  broke  off  below  tenon  at  south  end  ;  load 

dropped  1500  Ibs. 

North  header  virtually  broken  at  21833  Ibs.,  so  each  header 
bore  aifi*  =  10916  Ibs. 

The  tail-beam  which  did  not  break  was  heavy  and  full  of 
pitch,  and  of  a  coarser  grain  than  the  other. 

No.  86.  —  Test  of  yellow-pine  headers  by  means  of  tail-beams 
and  floor. 

The  headers,  6  inches  by  12  inches,  span  5  feet  4  inches, 


TRANSVERSE   STRENGTH  OF    TIMBER.  703 

were  mortised  at  each  end  into  the  trimmers  with  a  double 
tenon  and  joint-bolt.  The  trimmers  were  supported  on  jack- 
screws,  as  before.  Three  tail-beams,  3  inches  by  12  inches, 
span  10  feet,  were  mortised  into  the  headers  with  tusk  and 
tenon,  and  pinned. 

The  load  was  applied  at  the  centre  of  tail-beams,  and  dis- 
tributed equally  over  the  three  by  means  of  bridging.  (See 
No.  83.)  Tail-beams  cross-bridged  in  two  places  with  2-inch  by 
3-inch  spruce,  and  floored  over  with  i-inch  yellow-pine  flooring. 

Weight  of  tail-beams,  bridging,  flooring,  etc.,  which  was 
supported  by  headers  —  763  Ibs. 

No  deflections  taken. 

Weight  of  yoke  and  iron  bridging,  366  Ibs. 

Details  of  the  Test. 

14366  Ibs.     North  header  heard  to  crack  internally. 
24366    *'       Two  tail-pieces  began  to  crack  under  lower  tenon  (north 

end). 
24866    "       A  few  minutes  later  one  of  them  broke  under  tenon  (north 

end).     The  headers,  as   far   as   could   be   seen,  were 

uninjured. 

No.  89.  —  Test  of  headers  in  floor  (yellow  pine).  The 
headers  and  trimmers  were  the  ones  used  in  No.  86,  and  were 
framed  in  the  same  way.  Three  tail-beams,  3  inches  by  12 
inches,  6  feet  6  inches  span  (inside  measurement),  were  framed 
into  headers  with  a  tusk-and-tenon  joint,  and  then  pinned. 

The  experiment  was  precisely  the  same  as  No.  86,  with 
the  exception,  that,  instead  of  ro-foot  tail-beams  being  used,  the 
length  of  these  was  6  feet  6  inches. 

No  deflections  were  taken. 

Details  of  the  Test. 

*  20150  Ibs.     Season  cracks  in  north  header  began  to  open  wider. 

*  26325    "       North  header  broke  through  the   middle,  following  the 

line  of  mortises,  then  held  18825  Ibs. 
South  header  cracked  but  little. 

*  This  weight   includes   half   the  weight  of    bridging,  floor,  and  tail  beams 

(32  5  Ibs). 


7°4  APPLIED  MECHANICS. 

No.  105.  —  Tests  of  yellow-pine  header  in  floors. 

The  headers  and  tail-beams  were  framed  as  in  No.  86 ; 
and  the  experiment  was  exactly  the  same,  with  the  exception 
that  the  tail-beams  used  were  6  feet  long. 

Details  of  the  Test. 

20762  Ibs.     Season  crack  in  both  headers  opened. 
23262    "       North  header  failed,  after  three  minutes,  through  the  line  of 

mortises  ;  while  south  header  was  but  little  cracked. 
No.  106.  —  Test  of  yellow-pine  header. 

In  this  experiment  the  headers  were  hung  in  stirrup  irons, 
exactly  as  in  No.  83. 

Details  of  the  Test. 

24262  Ibs.     Slight  cracking. 

26662    "       East  tail-beam  split  below  line  of  tenon. 
30262    "       Held  for  five  minutes,  when  west  stirrup  on  north  tail-beam 
broke.     The  header  was  virtually  broken. 

No.  107.  —  Test  of  yellow-pine  header. 

The  unbroken  headers  of   Nos.  105  and  106  were  used. 

Details  of  the  Test. 

22262  Ibs.     Crack  in  north  header. 

25162    "       North  header  (framed)  suddenly  failed. 


TRANSVERSE   STRENGTH  OF   TIMBER. 


705 


ECagji 

I_K—   5x18 


i- 


(8) 


i 


TESTS    OF    SPRUCE    HEADERS. 

The  general  dimensions  of  the  floors  tested  are  shown  in 
the  figure.  In  all  these  tests  the  tail- 
beams  are  joined  to  the  headers  by  tusk 
and  tenon  joints.  The  load  was  dis- 
tributed equally  at  the  centres  of  the 
three  tail-beams. 

No.  141.  Spruce  floor  4"Xi2"  headers 
joined  to  trimmers  by  double  tenon  and  joint 
bolt.  Header  good  quality,  cracked  along 
centre. 


6839  Ibs**  cracks  begin  to  open  in  S  header. 

11429  "  tail-beam  No.  i  split  at  N  header,  load  fell  to  10339  lbs 

10839  "      "       "      No.  2         "     N      "          "        "  10539  " 

13239  "      "       "      No.  2        "      S       "          "        "  13029  " 

13239  "      "       "      No.  3         "    N       "          "        "  13039  " 

13039  "  S  header  split, • 11039  " 

14139  *'  tail-beam  No.  3  split  further  at  ^header,  load  fell  to  13439  '* 

18039  "  lower  half  of  south  header  broke  by  tension. 


No.  142.  Spruce  floor,  headers  hung  in  stirrups, 
are  straight-grained,  with  cracks  along  centre. 


Both  headers 


10739  Ibs.,  tail-beam  No.  2,  cracked  at  N  header,  load  fell  to  10439  lbs' 

14239    "       "       "      Nos.  i,  2,  3,  "      S      "  "        "  12729    " 

17139    "       "       "      No.  i,  "    N      "  "        "  16139    " 

20339    "       "       "      No,  3  broke  by  crushing  and  tension,  "         "  16439    " 

No  perceptible  injury  to  headers. 

Trial  No.  2  resulted  in  failure  of  tail-beams,Vith  no  perceptible 
injury  to  header. 

Trial  No.    3.    Yellow-pine   tail-beams.      Same   dimensions   as 
before. 

17839  Ibs.,  S  header  cracked  between  tail-beams  No.  Tand  trimmer. 
19839    "     S      "       split  along  middle,  load  fell  to  14639  Ibs. 

No.  143.  The  headers  are  the  successful  ones  of  the  two  pre- 
ceding tests.     The  S  header  is  joined  to  the  trimmer  by  double 


706  APPLIED   MECHANICS. 

tenon  and  joint  bolt,  and  is  somewhat  injured  by  previous  test.     N 
header  is  hung  in  stirrups. 

16839  Ibs.,  S  header  begins  to  crack. 

18829    "     S      "        broke  by  splitting,  load  fell  to  1839. 

No.  170.  Spruce  floor,  yellow-pine  tail-beams;  headers  3J-  X  12 
cross-section.  Good  quality  spruce,  season  cracks  along  centres. 

No.  1700.  Headers  joined  to  trimmers  with  double  tenon  and 
joint  bolt.  12833  Ibs.,  crack  opened  in  N  header  :  13833  Ibs., 
S  header  split  along  neutral  axis. 

No.  170^.  Same  as  «,  with  header  hung  in  stirrups,  substituted 
for  broken  one. 

7733  Ibs.,  crack  opening  in  S  header. 
14833    "     S  header  split  along  centre. 

No.  170*:.  Same  as  £,  with  header  hung  in  stirrups,  substituted 
for  broken  one. 

9033  Ibs.,  crack  beginning  to  open  in  S  header. 
19833    "     ^header  split  along  centre,  giving  way  completely. 
No.   217.  Dimensions  of  floor  altered  as   indicated   in   figure. 

Spruce  headers,  yellow-pine  tail-beam. 
%      n  *f  [  |         2\^A.   Headers  joined  to  trimmers  by  dou- 
ble tenon  and  joint  bolts.     Headers  4"Xi2" 
spruce,  clear  lumber,  straight-grained. 
14700  Ibs.,  S  header  began  to  crack. 
20000    "     S      "       split  along  centre,  giving 
way  completely. 


„ 

.>  ^n*"13  n  '15 

"11 


.  Headers  hung  in  stirrup  irons,  4"x  12"  spruce,  clear 
lumber,  straight-grained. 

27500  Ibs.,  centre  tail-beam  split  at  S  header,  load  fell  to  25500  Ibs. 
31700    "     S  header  split  along  centre,  giving  way  completely. 

2176*.  Unbroken  header  of  217^4  replacing  broken  header  of 
217.5.  S  header  joined  to  trimmers  by  double  tenon  and  joint 
bolt,  N  header  hung  in  stirrups. 

20900  Ibs.,  S  header  split  along  centre,  giving  way  completely. 


TRANSVERSE  STRENGTH  OF   TIMBER. 


707 


BAUSCHINGER'S  TESTS. 

In  the  ninth  Heft  of  the  Mittheilungen  are  given  the  results 
of  an  experimental  study  which  Bauschinger  made  of  the 
strength  of  certain  pine  and  spruce  woods,  in  connection  with 
their  other  properties,  as  specific-gravity,  age,  time  of  felling, 
etc. ;  but  special  attention  is  given  to  the  variation  of  strength 
and  specific  gravity,  with  the  percentage  of  moisture  which  they 
contain,  i.e.,  their  condition  of  dryness.  While  he  did  a  con- 
siderable amount  of  work  upon  the  variation  of  the  tensile 
strength  with  the  percentage  of  moisture,  the  results  are  rather 
variable,  and  none  of  this  work  will  be  quoted  here  for  the 
reasons  given  on  page646; under  Tension  of  Timber.  He  him- 
self came  to  the  conclusion  that  more  satisfactory  results  could 
be  reached  by  experimenting  upon  compressive  strength. 

The  following  tables  give  summaries  of  the  tests  which  he 
made  and  reported  in  this  ninth  Heft  upon  compressive  and 
transverse  strength : 

COMPRESSIVE    TESTS. 
TEST-PIECES  ABOUT  3^x3^  INCHES  AND  6  INCHES  LONG. 


Summer  Felled. 

Winter  Felled. 

Timber. 

Place. 

Percent- 

Compressive 

Percent- 

Compressive 

age  of 

Moisture. 

Strength. 
Mean  Values. 
Lbs.  per  Sq.  In. 

age  of 
Moisture. 

Strength. 
Mean  Values. 
Lbs.  per  Sq.  In. 

Pine    .     . 

Lichtenhof    . 

9 

3997 

26 

4537 

Spruce     . 

Frankenhofen 

20 

3449 

17 

4452 

Spruce     . 

Regenhiitte  . 

27 

3328 

20 

3997 

Spruce     . 

Schliersee     . 

20 

2304 

T9 

3200 

708 


APPLIED  MECHANICS. 


TRANSVERSE    TESTS. 


Per- 

Modulus of 

Modulus  of 

Timber. 

Place. 

Breadth  and 
Depth. 

centage 
of  Mois- 

Elasticity. 
Lbs.  per 

Rupture. 
Lbs.  per 

Inches. 

ture. 

Sq.  In. 

Sq.  In. 

I.  Summer  Felled.     Span,  8  ft.  2.43  ins. 

Pine    .     . 

Lichtenhof  .     . 

6.70X   6.8l 

24 

1422300 

6002 

« 

« 

7.I9X   7-17 

23 

1536084 

6685 

« 

« 

6.66X  6.72 

J9 

1536084 

6742 

it 

« 

7.o6X  7-17 

25 

1664091 

7453 

Spruce 

Frankenhofen  . 

6.72X  6.76 

29 

1706760 

6372 

« 

« 

7-32X  7-25 

35 

1649868 

6344 

«' 

<« 

7-77X  7-8o 

25 

1464969 

5703 

"         . 

« 

S.ioX  8.10 

26 

1436523 

5405 

« 

Regenhiitte  .     . 

7.52X  7-64 

39 

1578753 

6016 

« 

"          .     . 

8.oiX  8.07 

34 

1592976 

5476 

« 

« 

7.88X  7-83 

30 

1635645 

6173 

<  < 

ii 

8.36X  8.26 

32 

1720983 

6016 

if 

Schliersee    .     . 

10.17X10.24 

25 

1016945 

4025 

« 

« 

10.25X10.34 

24 

960052 

3840 

« 

ii 

10.47X10.47 

21 

1109394 

4281 

ii 

« 

10.73X10.67 

24 

1080948 

4281 

II.  Winter  Felled.     Span,  8  ft.  2.43  ins. 

Pine    .     . 

Lichtenhof  .     . 

6.58X  6.71 

33 

1422300 

6813 

it 

M 

6.23X  6.30 

33 

1664091 

7609 

11         .      . 

,        . 

7-igX  7-34 

3i 

1308516 

5348 

« 

tt 

8.30X  8.17 

34 

1479192 

5903 

Spruce     . 

Frankenhofen  . 

7.07X  6.94 

3i 

1479192 

5959 

" 

« 

7.i9X  7-29 

28 

1436523 

5903 

t( 

« 

6.65X  6.69 

24 

1863213 

6969 

« 

« 

6.64X  6.67 

24 

1820544 

6813 

it 

Regenhiitte  .     . 

6.38X  6.39 

27 

1479192 

6230 

ti 

ii 

7.02X  7-08 

29 

1536084 

6329 

lt 

ii 

6.66X  6.79 

30 

1635645 

6443 

ii 

n 

7.65X  7-66 

38 

1635645 

6358 

<« 

Schliersee    .     . 

10.98X12.66 

26 

1024056 

3641 

i« 

it 

10.83X13-18 

25 

981387 

3755 

ii 

€( 

11.19X11.23 

28 

967164 

3670 

« 

« 

ii.iiXn.io 

25 

981387 

3570 

TRANSVERSE   STRENGTH  OF   TIMBER.  709 

In  Heft  16  of  the  Mittheilungen,  Bauschinger  gives  an 
account  of  a  series  of  tests  of  the  crushing  and  transverse 
strength  of  the  more  important  coniferous  woods  from  the  dif- 
ferent districts  of  Bavaria. 

In  the  case  of  the  transverse  tests  the  percentage  of  moisture 
was  determined  by  experiment,  and  is  recorded  in  the  tables. 
Then  sections  were  cut  from  the  same  pieces  from  which  the 
beams  were  taken,  and  tested  for  crushing,  one  of  them  being 
rather  wet;  one  had  somewhere  near  15  per  cent  of  moisture, 
which  Bauschinger  considers  to  be  about  the  average  dryness 
of  the  air,  and  one  was  somewhat  drier ;  and  in  each  case  the 
percentage  of  moisture  is  determined  and  recorded. 

The  results  of  the  crushing  tests  are  then  plotted,  and  a 
curve  drawn,  from  which  he  determines  the  crushing-strength 
with  15  per  cent  of  moisture.  A  similar  proceeding  is  adopted 
m  regard  to  the  specific-gravity. 

The  45  sections  were  each  cut  into  five  specimens  about  3 
•or  4  inches  square,  one  of  them  containing  the  heart. 

These  (which  contain  the  heart)  he  omits  from  his  curves 
and  calculations,  and  plots  only  the  results  of  the  others. 

His  results  are  given  in  the  following  table,  a  perusal  of 
which  will  show  that  the  moduli  of  rupture,  and  also  the  crush- 
ing-strengths, run  somewhat  higher  than  they  do  for  woods  of 
the  same  name  in  the  tests  made  at  the  Massachusetts  Institute 
of  Technology.  This,  of  course,  maybe  due  to  the  woods  that 
Bauschinger  tested  being  stronger  than  American  woods  of  the 
same  name,  but  it  is  more  probably  due  to  the  facts  that,  i°,  the 
specimens  he  used  were  rather  smaller,  and,  2°,  they  were,  on 
the  whole,  drier  than  the  American  woods  tested  at  the  Mas- 
sachusetts Institute  of  Technology. 


710 


APPLI&D  MECHANICS. 


fc 

Mean  Compressive 

M 

Strength  of 

Transverse  Test. 

O 

Pieces  not  con- 

Span, 8  ft.  2.42  ins. 

•I 

taining  Heart, 

1 

reduced  to 

1 

V) 

Place  of 

a 

b 

•R 

1 

Name. 

Growth. 

cific  Gravity 
of  Moistute. 

15* 
Moisture. 
Lbs.  per 
Sq.  In. 

Same 
percent 
Moisture 
as  in 
Trans- 
verse 
Test. 

11 

centage  of  M( 
ture. 

Modu- 
lus of 
Elas- 
ticity. 
Lbs.  per 
Sq.  In. 

Modu- 
lus of 
Rup- 
ture. 
Lbs. 

6 

| 

Lbs.  per 

s*« 

I 

Sq.In. 

* 

CO 

Sq.  In. 

CQM 

Q- 

£ 

i 

Larch 

St.  Zeno 

0.62 

6827 

6756 

5.78 

6.97 

15-5 

2076560 

10382 

2 

Larch 

0.67 

7353 

6784 

5.85 

6.99 

17.2 

1692540 

10596 

3 

Pine 

11 

°-53 

6045 

6400 

5-78 

5-74 

13-5 

1834770 

9742 

4 

Spruce 

1 

0-43 

4978 

4551 

7-73 

7.61 

16.6 

1479190 

7183 

5 

Pine 

' 

0.52 

5263 

5120 

7-75 

7-7° 

IS  8 

1649888 

7325 

6 

Spruce 

4 

0-45 

5405 

5334 

7.69 

9.21 

1635650 

6756 

7 

44 

4 

0.48 

5547 

5831 

5-80 

6.85 

J4-3 

1592980 

7965 

8 

" 

* 

0.51 

5974 

5689 

6.87 

6.82 

16.1 

^7875° 

9 

Larch 

1 

0.59 

6685 

4978 

6.10 

6.91 

16.0 

1706760 

9245 

10 

" 

" 

0.58 

6116 

5974 

5-85 

6.91 

15.5 

1521860 

9743 

ii 

12 

Zlirbe 
Larch 

Karlstein 

0.41 

O.OI 

3200 
796S 

3271 
'7752 

5-86 
5-85 

7.00 
7.16 

14.7 

15-8 

846269 
2097898 

9529 

13 
14 

Spruce 
Pine 

a 

0.54 

0-54 

5974 
5974 

5334 
6258 

6.79 
5-65 

7.78 
6-93 

17.4 
14.4 

1592980 
1905880 

7965 
10027 

Spruce 

" 

0.46 

54°5 

5I91 

6.78 

7.80 

16.3 

1777880 

8107 

16 

Larch 

" 

0.68 

7965 

7325 

7.61 

9-34 

17.9 

2039670 

8605 

17 

Spruce 

*•        ,  - 

0-54 

6969 

6898 

5-79 

6-95 

15.1 

2026780 

9387 

18 

'• 

«(        —  - 

0.49 

6045 

5476 

5-86 

7.80 

18.3 

1578750 

6542 

19 

" 

*' 

0.52 

6187 

6045 

5-83 

6.87 

15-6 

1692540 

8036 

20 

Pine 

ii 

0.52 

5831 

59°3 

5-77 

7.71 

14.9 

I55°3°7 

7823 

21 
22 

Larch 
Spruce 

it 

0.61 
0.51 

7040 
6471 

6258 
6187 

5.81 

6-93 
7-77 

17.6 
15-8 

1905880 

10241 
8107 

23 
24 

Larch 

M 

Freising 

°-39 
0.61 

4267 
7823 

4°54 
7538 

7.72 

5-75 

9-31 

6.94 

15-9 

1208960 
2062340 

5334 
9814 

25 

44 

0.70 

8249 

7538 

5.84 

6.91 

16.9 

2176120 

9814 

26 

Spruce 

1 

0.59 

6969 

5831 

6.87 

7.72 

18.4 

1977000 

8249 

27 

44 

* 

0.44 

5405 

5191 

6.96 

7.62 

15-6 

1479190 

6685 

28 

Pine 

4 

0.52 

6187 

5476 

6.88 

7-75 

18.4 

6258 

29 

44 

4 

0.62 

7894 

7467 

6.86 

7.64 

16.9 

2062340 

9814 

30 

31 

Spruce 

i 

0.48 
0.47 

SJS 

5689 
4480 

6.91 
6-93 

7-74 
7.80 

16.7 
18.6 

1607200 
1763650 

7609 
6898 

32 

Pme 

' 

0-53 

6116 

5689 

5-83 

6.86 

16.8 

1848990 

8818 

33 
34 

Spruce 

! 

0.49 
°-59 

4907 
6827 

6400 

ill 

6.91 
7-74 

17.0 
16.7 

1223180 
1948550 

5974 
9103 

35 

44 

*  • 

0.48 

5618 

5334 

6.96 

7.70 

16.4 

1678310 

647^ 

36 

44 

• 

0.44 

5334 

5I91 

6.96 

7.71 

15-8 

1550310 

7325 

37 

44 

* 

0.49 

6258 

5974 

7.00 

7.82 

16.4 

1806340 

7467 

38 

Larch 

" 

0.58 

7325 

7112 

5-77 

6.91 

iS-5 

1742320 

9387 

39 

14 

44 

0.65 

59°3 

5263 

7.00 

6.96 

18.2 

1550310 

7752 

40 
4r 

Pine 
Spruce 

M 

0.49 
0.46 

4764 
6116 

4267 
5547 

6.96 
6.88 

7-81 
7.80 

17.7 
17.0 

1038280 
1280070 

3485 
5618 

42 

44 

Unterliezheim 

0.42 

4120 

4125 

6.85 

7.71 

20.7 

1464970 

6i44 

43 

44 

44 

0-39 

4907 

4551 

6-95 

7.80 

18.8 

1251620 

5^91 

44 

White  Pine 

" 

0-33 

34H 

3058 

6.99 

7.83 

18.1 

810731 

4I25 

45 

if        ii 

0.32 

3200 

2631 

7.00 

7-84 

21.4 

72537° 

3556 

TRANSVERSE   STRENGTH  OF   TIMBER. 


711 


AVERAGE   COMPRESSIVE    STRENGTH    OF   WHOLE    SECTION   OF    LOG. 

POUNDS  PER  SQUARE  INCH. 


Pine  from 

Spruce  from 

Spruce  from 

Spruce  from 

Time  of 

Lichtenhof. 

Frankenhofen. 

Regenhiitte. 

Schliersee. 

i 

2 

• 

2 

i 

2 

i 

2 

Summer 

7183 

5244 

6416 

4807 

6287 

5319 

4570 

3143 

Winter 

6343 

6784 

6756 

56l8 

6343 

5348 

4779 

4238 

i.  Tested  5  years  after  felling. 


2.  Tested  3  months  after  felling. 


§240.  Shearing  of  Timber  along  the  Grain.  —  The  shear- 
ing of  timber  always  takes  place  along,  and  not  across,  the 
grain ;  for  it  can  be  shown,  that,  wherever  we  have  a  tendency 
to  shear  on  a  certain  plane,  there  is  an  equal  tendency  to  shear 
on  a  plane  at  right  angles  to  it.  Hence  if  there  is,  at  any  point 
in  a  piece  of  wood,  a  tendency  to  shear  it  across  the  grain,  there 
must  necessarily  accompany  it  an  equal  tendency  to  shear  it 
along  the  grain  ;  and,  the  resistance  to  the  latter  being  very 
slight,  the  timber  will  give  way  in  this  manner,  instead  of  across 
the  grain. 

As  to  the  shearing-strength  per  square  inch,  some  values 
have  been  given  in  Rankine's  table ;  and  the  following  table 
contains  results  obtained  at  the  Watertown  Arsenal,  and  re- 
corded in  Executive  Document  No.  12,  47th  Congress,  first 
session. 


712 


APPLIED   MECHANICS. 


Kind  of  Wood. 

Arsenal 
No. 

Shearing- 
Strength 
per  Square 
Inch. 

Kind  of  Wood. 

Arsenal 
No. 

Shearing- 
Strength 
per  Square 
Inch. 

Ash    

62O 

600 

Oak  (white)     .    . 

631 

7^2 

621 

592 

Pine  (white)     .    . 

*J 

752 

/  J 

324 

622 

458 

753 

267 

623 

700 

754 

352 

Birch  (yellow)      . 

623 

563 

755 

366 

* 

633 

8I5 

Pine  (yellow)  .    . 

607 

399 

634 

672 

608 

3*7 

635 

612 

614 

409 

Maple  (white)  .    . 

636 

647 

615 

415 

637 

537 

616 

409 

638 

367 

617 

364 

639 

43i 

618 

286 

Oak  (red)    .    .    . 

624 

775 

619 

330 

625 

743 

Spruce    .... 

748 

253 

626 

999 

749 

374 

627 

726 

75° 

347 

Oak  (white)     .    . 

628 

966 

75i 

3i6 

629 

803 

Whitewood     .    . 

609 

406 

630 

846 

610 

382 

§  241.  General  Remarks.  —  A  perusal  of  the  tests  on 
columns  and  on  beams  will  show  that  one  of  the  principal 
sources  of  weakness  in  timber  is  the  presence  of  knots,  and  it 
will  be  noticed  that  the  position  of  the  fracture  is  in  most 
cases  determined  by  the  knots. 

Sap-wood,  season  cracks,  and  decay  are  doubtless  other 
sources  of  weakness.  The  tests,  however,  do  not  present  such 
striking  evidence  of  the  deleterious  effects  of  the  first  two  as 
is  the  case  with  knots.  In  general,  it  may  be  said,  however, 
that  timber  used  in  construction  should  be  free,  or  nearly  free, 
from  sap-wood ;  as  an  excessive  amount  of  sap-wood  renders  it 
weak. 


GENERAL   REMARKS.  713 


It  will  often  be  found  to  be  a  common  opinion  among  lum- 
ber-dealers, that  a  piece  of  timber  which  contains  the  heart  is 
not  as  good  as  one  which  is  cut  from  the  wood  on  one  side  of 
the  heart.  This  is  very  often  true  ;  as  the  timber  which  is  sold 
in  the  market  is  very  liable  to  have  cracks  at  the  heart,  and 
also,  if  the  tree  has  passed  maturity,  the  heart  is  the  place 
where  decay  is  likely  to  begin.  Nevertheless,  the  tests  of 
beams  would  not,  it  seems  to  the  author,  bear  out  the  conclu- 
sion that  such  pieces  as  contain  the  heart  are  always  weaker 
-than  those  that  do  not. 

Another  matter  that  claims  serious  consideration  is  the 
effect  of  seasoning  upon  the  strength  of  timber.  This  question 
can  only  be  decided  by  tests  on  full-size  pieces,  as  the  small 
pieces  season  much  more  rapidly  and  uniformly  than  full-size 
pieces. 

In  this  regard,  the  observation  should  be  made,  that  prac- 
tically our  buildings  and  other  constructions  are  built  with 
green  lumber ;  i.e.,  lumber  which  has  been  cut  from  three 
months  to  a  year.  Unless  it  can  be  shown  that  the  seasoning 
which  the  lumber  receives  while  in  use  imparts  to  it  a  greater 
strength,  it  will  only  be  proper  to  consider  its  strength  the  same 
as  that  of  green  lumber.  Not  very  much  evidence  has  thus  far 
been  obtained  upon  this  point ;  but,  such  as  it  is,  it  will  be 
noted  here. 

i°.  We  have,  on  p.  505,  the  results  of  the  tests  of  a  lot  of 
old  mill  columns  ;  and,  while  some  of  them  did  exhibit  a  greater 
strength  than  green  ones,  a  perusal  of  this  set  of  tests  will 
convince  the  reader  that  it  would  not  be  safe  to  rely  upon  any 
greater  strength  in  these  columns  than  in  green  ones.  More- 
over, these  columns  had  been  in  a  building  heated  by  steam  for 
a  number  of  years,  and  during  the  seasoning  process  they  had 
been  subjected  to  the  load  they  had  to  support.  The  writer 
has  also  observed  some  evidence  of  the  same  kind  in  con- 
nection with  one  of  his  time  tests. 


714  APPLIED   MECHANICS. 

2°.  In  the  case  of  beams,  we  have,  in  Nos.  60  and  66, 
examples  of  beams  which  had  been  seasoning,  unloaded,  in  a 
building  heated  by  steam  ;  and  in  these  cases  there  was  a  great 
gain  in  strength.  Some  yellow-pine  beams  exhibited  a  similar 
action.  On  the  other  hand,  beams  Nos.  18  and  19  had  been 
seasoning  on  the  wharf,  in  the  open  air,  for  about  one  year ; 
and  while  some  yellow-pine  beams  which  had  seasoned  without 
load,  in  the  building,  showed  great  strength,  in  other  cases  the 
increase  was  not  so  marked. 

In  view  of  the  fact  that  the  above  is  practically  all  the  evi- 
dence we  have  in  the  matter,  it  would  seem  to  the  writer,, 
unless  future  experiments  shall  prove  the  contrary  to  be  true, 
that  we  cannot  rely,  in  our  constructions,  upon  having  any 
greater  strength  than  that  of  the  green  lumber,  and  that  the 
figures  to  be  used  should  be  those  obtained  by  testing  green 
lumber. 

§  242.  Building-Stones. — The  three  most  important  factors 
about  a  building-stone  besides  its  beauty,  are  its  durability,  its 
strength,  and  the  ease  with  which  it  can  be  quarried  and 
worked.  In  order  to  be  durable  it  must  be  able  to  withstand 
the  deleterious  influences  of  rain,  wind,  frost,  fire,  and  of  the 
acids  that  are  found  in  the  air  especially  in  large  cities,  where 
the  most  common  are  carbonic  acid  and  sulphur  acids. 

The  durability  of  a  stone  is  probably  its  most  important 
feature,  and  is,  perhaps,  the  most  difficult  to  test  thoroughly. 
As  a  rule,  the  greater  -its  hardness  and  the  less  its  absorptive 
power  for  moisture,  the  more  durable  will  it  be. 

Tests  of  hardness  are  easily  and  frequently  made.  Tests  of 
absorptive  power  for  moisture  are  very  often  made.  While 
the  methods  pursued  by  different  people  differ,  they  consist 
essentially  of  weighing  the  specimen  dry,  and  then  soaking  it 
in  either  hot  or  cold  water  until  it  has  absorbed  all  that  it  will,, 
and  weighing  it  again. 

There  are  a  number  of  methods  pursued  in  order  to  de- 


B  UILDING-STONES.  J 1 5 


termine  its  power  of  withstanding  the  action  of  frost,  and  the 
results  differ,  of  course,  according  to  the  method  pursued. 
Bauschinger's  method  consists  in — 

i°.  Determining  the  compressive  strength  of  the  stone  in 
a  dry  and  in  a  saturated  condition,  and  comparing  the  two. 

2°.  Determining  the  compressive  strength  after  twenty- 
five  freezings  and  thawings. 

3°.  Determining  the  loss  of  weight  after  these  twenty-five 
freezings.  . 

4°.  Examining  the  specimen  with  a  microscope  for  cracks 
after  the  twenty-five  freezings. 

Stones  will  not  withstand  the  heat  of  a  large  conflagration, 
brick  being  better  than  any  building-stone. 

As  to  how  well  a  stone  will  stand  the  gases  in  the  air  of  a 
large  city,  a  great  many  tests  have  been  proposed  and  used,  but 
none  of  them  are  entirely  satisfactory.  Of  course  we  can  get 
indications  from  a  study  of  the  chemical  composition,  or  better 
from  a  microscopical  examination,  which  shows  also  the 
arrangement  of  the  different  components,  this  being,'  of  course, 
the  part  of  the  geologist  or  mineralogist. 

After  the  question  of  durability,  the  strength  comes  in  as 
the  factor  of  next  importance,  and  although  the  loads  usually 
put  upon  stones  in  construction  are  very  much  smaller  than 
the  breaking-strength  of  the  stone  as  shown  by  small  speci- 
mens tested  in  the  testing-machine,  nevertheless  the  mortar 
or  cement  joints,  the  bonding,  and  the  necessary  unevenness 
render  the  real  factor  of  safety  very  much  less  than  would  be 
at  first  imagined. 

Building-stones  are  more  often  called  upon  to  bear  a 
compressive  load  than  any  other,  though  they  are  sometimes 
called  upon  to  bear  a  transverse  load,  as  in  the  case  of  window- 
lintels. 

The  results  are  very  variable,  partly  because  the  stone 
varies  very  much  in  quality,  and  partly  because  it  is  only  of 


716  APPLIED  MECHANICS. 

late  years  that  it  has  been  recognized  that  in  order  to  obtain 
correct  results  in  compression  tests  the  pressure  must  be  evenly 
distributed  over  the  surfaces  of  the  specimen  pressed  upon, 
and  that  in  order  to  accomplish  this  even  distribution  it  is 
necessary  that  the  faces  which  come  in  contact  with  the  plat- 
forms of  the  testing-machine  shall  be  accurate  planes,  and  that 
unless  the  compression  platforms  are  adjustable,  the  two  faces 
pressed  upon  shall  be  parallel — provided,  of  course,  the  plat- 
forms are  parallel,  as  they  should  be.  Formerly  it  was  thought 
that  the  desired  result  could  be  obtained  by  interposing  be- 
tween the  surface  of  the  specimen  and  the  platform  some 
soft  substance,  as  a  cushion  of  wood  or  of  lead,  whereas  it  is  a 
fact  that  any  such  cushions  only  render  the  results  smaller 
and  more  variable  than  the  real  crushing-strength  of  the  speci- 
mens. Hence  it  is  that  a  great  many  of  the  tests  that  have 
been  made  are  of  no  value,  because  this  matter  was  not 
attended  to.  In  a  rough  way  we  may  divide  the  most  com- 
mon building-stones  as  follows  :  i°.  Granites  and  allied  stones  ; 
2°.  Limestones,  including  marbles ;  3°.  Sandstones ;  4°.  Slates. 

The  most  systematic  set  of  tests  was  made  by  Bauschinger, 
and  is  reported  in  the  Mittheilungen,  Hefte  i,  4,  5,  6,  10,  n, 
1 8,  and  19. 

Besides  this  we  may  n  ote  the  following  references  : 

i°.  Report  on  Compressive  Strength,  etc.,  of  the  Building-Stones  in 

the  United  States,  1876.     Gillmore. 
2°.  Compressive  Resistance    of   Freestone,  Brick  Piers,  Hydraulic 

Cements,  Mortars,  and  Concretes,  1888.     Gillmore. 
3°.  Masonry  Construction,  1889.     Baker. 
4°.  Testing  Materials  of  Construction,  1888.     Unwin. 
5°.  History  of  the  St.  Louis  Bridge,  1881.     Woodward. 
6°.  Exec.  Doc.  12,  47th  Congress,  ist  session.     Senate. 
7°.  Exec.  Doc.  35,  4Qth  Congress,  ist  session.     Senate. 

Some  tables  of  results  will  now  be  given. 


B  UILDING-STONES. 


717 


The  following  is  taken  from  Woodward's  "  History  of  the 
St.  Louis  Bridge :" 


Material. 

Length, 
in 
inches. 

Diameter  of 
Cross-section, 
in  inches. 

Modulus  of 
Elasticity, 
pounds  per 
square  inch. 

Breaking- 
Weight,  per 
square  inch. 

Grafton  Magnesian  Limestone. 
«                 K                 « 

u                it                (t 
ti                tt                 tt 

"                "      (5  specimens) 

6.46 
5.87 
5.96 
5-99 
3.00 

1.  1.4 
1.  06 
1.  06 
1.07 
3  X  3 

10500000 
8400000 
8500000 
6000000 

7200 
8500 
2000 
6000 
av.  154.00 

tt                tt 
tt                tt 

Portland  Granite  

8.00 
13.00 

5  88 

2.38 
I.J3 

2    16 

12000000 
5000000 
e  COOOOO 

IOIOO 

10800 
16000 

tt               ti 

«?  08 

2.^6 

64OOOOO 

18500 

tt               n 

5Q7 

2    18 

5OOOOOO 

17000 

Richmond  Granite     .  • 

6  oo 

o    ao 

I35OOOOO 

16400 

<!     OO 

q   V   q 

I37OO 

Missouri  Red  Granite  

3.00 

3  X  3 

I27OO 

tt                     tt 

a    OO 

•?  x  T 

I3OOO 

tt                     n 

•3.    OO 

•3    V    3 

I27OO 

it                     tt 

300 

o  y  «i 

13600 

Brown  Ochre  Marble  

3  .00 

•i  V  ^ 

I5OOO 

Sandstone,  St.  Genevieve,  Mo. 

300 

•3    V    -3 

eqao 

«                        tt 

4  88 

4  88  X  4  88 

eeoo 

tt                 i      tt 

3  06 

3  06  X  3  06 

34.OO 

In  Heft  4  of  the  Mittheilungen,  Bauschinger  gives  a  long 
table  of  results  of  testing  granites,  limestones,  and  sandstones, 
from  which  the  following  examples  are  selected : 


APPLIED  MECHANICS. 


Kind  of  Stone. 

Place. 

Tests  by  Compres- 
sion. 

Transverse  Tests. 

Crushing 
Strength, 
pounds 
per  sq.  in. 

Direction 
of  the 
Pressure 
with  re- 
spect to 
the  bed. 

Modulus 
of  Rup- 
ture, 
pounds 
per  sq.  in. 

Direction 
of  Break- 
ing Sec- 
tion with 
respect  to 
the  bed. 

Granites. 
Yellow,  very  soft,  exceptional  quality 

Gray    coarse-grained    .    .         ... 

Selb  in  Ober- 
f  ran  ken. 

M 

Waldstein 
St.  Gotthard 

u 

Cham 
Ebendaher 

Schlanders  in 
Tyrol 
Wurzburg 
Kronach 

(4 

Kelheim 

7750 
11300 
11720 

14790 
11740 
12660 
15650 
13230 

22190 
19200 

12800 
6260 
22760 
11390 

\     6900 
1     8390 
3|6o 
(    10600 
•{   20340 
(    17920 

j    18490 
j    16780 
j    12520 
(    11240 

4836 

19270 
20550 

11950 
8100 
j     8620 
1      939° 
654C 

2420 

2490 
4340 

5480 
2680 

j      44io 
1      3630 

1 
1 
II 

1 
1 

1 
II 

1 

1 

1 
II 

1 

1 
II 

1 

f 

1 
II 

}  x 

II 

1 

1 
1 

1 

J. 

1 
II 

av.i94o 

1309 
2773 

1991 

980 

1252 
939 

j-  2560 

oblique 
II 
across 

u  '4  *  . 

Very  light-colored,  tolerably  coarse- 
grained         
Very  hard,  coarse-grained    .... 

Very  hard,  striped    

White,    very    hard,    tolerably  fine- 
grained, only  good  for  paving  . 
Syenite,  black,  with  a  good  deal  of 

Limestones. 
White  marble                          . 

Muschelkalk 

u 

Yellowish  white,  soft  limestone  of  the 

Best  quality           

Poorest  quality.    ,    

Rosenheim 

Poppenheim 
Herzbruch 

Kronach 
Unterfranken 

Carlsruhe 
Wurtemburg 

Ansbach 

M 

Nuremberg 
Regensberg 

Kelheim 

Dolomite. 
From  the  white  Frankenjura.    .    .    . 
u              ik               d 

Sandstone. 
Bunter^sandstone,  gray,  with  yellow 

Bunter  sandstone,  red,   very  rich  in 

Do.              do.              do. 
Bunter    sandstone,  dark    red,    fine- 
grained   

Do.              do.              do. 
Keuper  sandstone,  red,  fine-grained. 

Do.              do.              do. 
Keuper  sandstone,  white,  with  red- 
dish bed  stripes,  coarse-grained. 
Keuper    sandstone,  white,   tolerably 
fine-grained  

Keuper  sandstone,  brown      .... 
Green  sandstone,  yellow,  with  brown 

Green  sandstone,  dirty  green,  tolera- 

Green     sandstone,     greenish,    fine- 

BUILDING-STONES.  719 


In  Heft  5  of  the  Mittheilungen  is  to  be  found  a  study  of 
the  modulus  of  elasticity  of  building-stones.  Bauschinger 
found  that  in  this  case  the  departure  from  Hooke's  law  is 
greater  with  small  than  with  large  loads.  Heft  6  contains  an 
experimental  study  of  the  laws  of  compression.  Heft  10  con- 
tains an  experimental  investigation  of  the  principal  Bavarian 
building-stones.  Hefte  11  and  18  contain  a  study  of  the  com- 
pressive  strength  and  wearing  qualities  of  paving-stones.  Heft 
19  contains  a  study  of  the  power  of  stones  to  resist  frost.  For 
all  these  the  student  is  referred  to  the  Mittheilungen. 

In  Exec.  Doc.  35,  49th  Congress,  ist  session,  Senate,  are  to 
be  found  the  results  of  a  set  of  tests  of  the  compressive  strength 
of  Haverstraw  freestone  cubes.  The  following  account  of 
the  tests  is  compiled  from  the  document  referred  to : — 

"  These  specimens  were  taken  from  a  quarry  at  Haver- 
straw,  N.  Y.  The  quarry  had  not  been  worked  for  some  time, 
but  was  reopened  for  the  purpose  of  obtaining  these  blocks, 
this  stone  being  selected  from  its  supposed  uniformity  .in 
strength  and  low  ultimate  resistance,  on  account  of  which  the 
series  could  be  carried  out  in  larger  blocks  and  not  exceed  the 
capacity  of  the  testing-machine.  The  series  includes  cubes 
from  i  inch  to  12  inches,  advancing  by  inches,  and  the  follow- 
ing prisms  : 

i"  X  i"  X  2",  4"  X  4"  X  i",  4"  X  4"  X  2",  4"  X  4"  X  3", 

8"  X  8"  X  2",  X  3",  X  4",  X  sV  X  6",  x;". 
The  bed  surfaces  were  faced  with  plaster  of  Paris,  and  tested 
between  steel  cushions." 

The  weight  per  cubic  foot,  of  the  different  samples,  varied 
from  129.2  Ibs.  to  140.7  Ibs.  The  results  are  given  in  the  fol- 
lowing tables  : 


720 


APPLIED  MECHANICS. 


COMPRESSIVE    STRENGTH    OF   CUBES   OF    HAVERSTRAW    FREESTONE.. 


Number 
of  Cubes. 

Nominal 
Dimensions 
in  Inches. 

Average  Ulti- 
mate Strength 
in  Ibs.  per 
Sq.  In. 

Remarks. 

2 

I 

7033 

4 

2 

6004 

4 

3 

6245 

4 

4 

5961 

4 

5 

6468 

4 

6 

7355 

4 

7 

6156 

4 

8 

6266 

4 

9 

6535 

3 

10 

6100 

The  fourth  specimen  stood 

8032    Ibs.    per    sq.    in. 

without  breaking. 

4 

ii 

6428 

i 

12 

5323 

The  other  three  stood  on 

the  average  5585  Ibs.  per 

sq.  in.  without  breaking. 

COMPRESSIVE    STRENGTH    OF    PRISMS    OF    HAVERSTRAW    FREESTONE. 


Nominal 

Average  Ulti- 

Number 
of  Prisms. 

Dimensions 
of  Prisms 

mate  Strength 
in  Ibs.  per 

Remarks. 

in  Inches. 

Sq.  In. 

2 

4X4X1 

16428 

2 

4X4X2 

8029 

2 

4X4X3 

6679 

2 

8X8X2 

12455 

Prisms  did    not  fracture  ; 

sides  flaked  off. 

2 

8X8X3 

9537 

2 

8X8X4 

8550 

2 

8X8X5 

7764 

2 

8X8X6 

6535 

2 

8X8X7 

6613 

I 

I  X  I  X  2 

4213 

3  UILDING-  S  TONES. 


721 


These  tests  are  also  published  in  Gillmore's  book  of  1888, 
and  they  disprove  the  position  he  took  in  1876,  that  the  com- 
pressive  strength  per  square  inch  increases  with  the  size  of  the 
specimen. 

In  the  same  Exec.  Doc.  is  given  a  set  of  tests  of  three  two- 
inch  cubes  of  Kanawha  freestone,  the  average  crushing- 
strength  being  9154  Ibs.  per  square  inch,  and  the  average 
weight  per  cubic  foot  151.9  Ibs. 

The  following  table  is  taken  from  the  Trans.  Am.  Soc.  Civ. 
Eng.  for  Oct.  1886,  where  it 'is  quoted  from  "  Mechanical  tests 
of  building  materials,  made  at  the  Watertown  Arsenal,  by  the 
United  States  Ordnance  Dept.,  at  the  request  of  the  Commis- 
sioners for  the  erection  of  the  Philadelphia  Public  Buildings:" 


Kind  of 
Stone. 

Locality. 

Color. 

Direction  of 
Pressure. 

Total 
Load 
applied, 
Ibs. 

Crushing 
Strength 
per  sq.  in. 
applied, 
Ibs. 

"rt    ^   . 

Remarks. 

r 

Lee,  Mass  

Blue. 

End. 

715000 

20504 

34.87 

Burst  in  fragments. 

'                  .... 

.  White. 

Bed. 

800000 

22370 

35  -16 

Slight  flaking. 

'                   .... 

W.  &B. 

End. 

800000 

22860 

34-99 

No  apparent  injury. 

'                  .... 

White. 

" 

800000 

22820 

35-05 

11                    II                                11 

'                  .... 

Blue. 

Bed. 

800000 

22900 

34-93 

Flaked,  one  edge. 

1 

.... 

W.  &B. 

" 

767000 

21700 

35-34 

Crushed  suddenly. 

ill 

Montgomery  Co.,  Pa. 

Blue. 

" 

466300 

11470 

40  64 

Failed  suddenly. 

5 

u 

4 

End. 

400000 

10420 

38.40 

Ultimate  strength. 

"                     " 

4 

Bed. 

543000 

13700 

39-63 

u 

"                     " 

4 

End. 

398000 

IOI20 

39-33 

U                              1, 

ii           .         ii 

' 

" 

3475oo 

9590 

36.24 

11                             11 

ii                     i* 

4 

Bed. 

434000 

10940 

39-67 

II             ,1 

r 

Conshohocken,  Pa.    . 



End. 

494000 

14090 

35-05 

Ultimate  strength. 

4) 

it                           it, 

Bed. 

cfifiooo 

ft      AC\ 

f- 

ii              ii 

G 

Indiana 

End. 

ft 

34  -°3 

n               u 

377000 

53° 

44  •  22 

i.              11 

3 

'.'.'.'.'. 



Bed. 

320500 
321000 

7190 
7776 

44  •  5^ 

41-38 

.i 

-i--';   .    .   . 



" 

438300 

10620 

41.28 

.1              n 

~* 

Vermont   J 

Dove-     ( 

Bed. 

' 

, 

Ult'       t                th 

rt  i 

colored  ' 

End. 

53     oo 
770800 

3400 

39  °5 
38  4.8 

C.i 

^/youu 

9 

/22 


APPLIED   MECHANICS. 


•sg 

II 

« 

Locality. 

Color. 

Direction  of 
Pressure. 

Total 
Load 
applied, 
Ibs. 

Crushing 
Strength 
per  sq.  in. 
applied, 
ibs. 

"H   -e 

Is* 

£<C/3 
C/) 

Remarks. 

r 

Hummelstown,  Pa.   . 

Bed. 

528700 

12810 

41.28 

Ultimate  strength. 

Ohio      

Buff. 

End. 
Bed. 

570300 

13610 
6510 

41.92 

39  -32 

End. 

4860 

41  .02 

i               n 

« 

„ 

t 

Bed 

7020 

41  -25 

i                u 

o 

, 

4 

End. 

394° 

4.0.06 

Bear  ngs  imperfect 

£1 

Blue. 

7680 

39.68 

Ultimate  strength. 

c 
rt 

( 

Bed 

41  .go 

1                                    U 

C/J 

, 

, 

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40.00 

(                                U 

, 

, 

Bed 

SylO 

I                                   U 

, 

tt 

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41  28 

Fractured  suddenly. 

, 

u 

Bed 

Ultimate  strength 

Bauschinger,  in  Heft  12  of  the  Mittheilungen,  gives  an  ac- 
count of  fifteen  tests  on  the  effect  of  heat,  and  sprinkling  with 
water,  on  the  following  kinds  of  posts : 
i°.  One  gray  coarse-grained  granite. 
2°.  One  red  marble. 
3°.  One  tuffa. 
4°.  One  dolomite. 

One  granitic  marble. 
Three  Bunter  sandstone. 
Two  green  sandstone. 
Two  Keuper  sandstone. 
One  concrete. 
.  Two  brick. 

The  results  of  these  tests  showed  that  the  concrete  and  the 
brick  posts  stood  the  fire  and  sprinkling  quite  well ;  but  none 
of  the  stone  posts  were  able  to  resist  such  treatment. 

§  243.  Hydraulic  Cements  and  Brick  Piers. — When  a 
pure  or  nearly  pure  limestone  is  calcined,  so  as  to  drive  off  the 
carbonic  acid,  we  have,  as  a  result,  an  oxide  of  lime,  commonly 
called  quick-lime,  which  has  a  great  avidity  for  water,  and 


5° 
6°. 


9- 

ioc 


HYDRAULIC  CEMENTS  AND  BRICK  PIERS.  723 

which,  on  the  addition  of  water,  slakes,  with  the  development  of 
a  considerable  quantity  of  heat ;  and  the  result  is  the  formation 
of  a  hydrate,  in  the  form  of  a  fine  powder,  which  by  the  addi- 
tion of  more  water  is  reduced  to  a  paste,  which  slowly  hardens 
upon  exposure  to  the  air.  It  is  this  paste,  mixed  with  sand, 
which  forms  the  mortar  so  extensively  used  in  ordinary  cheap 
buildings.  It  is  very  weak,  and  hardens  very  slowly,  even  in 
the  air. 

On  the  other  hand,  a  hydraulic  lime  or  a  hydraulic  cement 
contains  a  certain  amount  of  impurities,  of  which  silica  forms 
the  principal  portion,  though  we  usually  find  also  alumina,  pro- 
toxide of  iron,  and  magnesia ;  and  these  impurities  are  contained 
in  so  large  a  proportion  that  the  slaking  entirely  or  nearly  dis- 
appears, and  instead,  the  addition  of  water  after  calcination 
causes  the  formation  of  hydrated  silicates,  etc.,  which  harden 
as  well,  or  better,  under  water  as  in  the  air. 

While  we  cannot  draw  a  sharp  line  of  demarcation  between 
hydraulic  lime  and  hydraulic  cement,  nevertheless  the  essen- 
tial difference  is  that  the  first  contains  pure  lime  in  sufficient 
proportions  to  slake,  but  at  the  same  time  contains  enough 
clay,  silica,  etc.,  to  enable  it  to  set  under  water ;  whereas 
hydraulic  cement  contains  less  pure  lime,  and  hardly  slakes  at 
all,  but  sets  more  rapidly  than  hydraulic  lime. 

Hydraulic  cements  are  known  as,  i°,  Portland  ;  2°,  Rosen- 
dale  ;  3°,  Roman  cement.  The  last  is  an  Italian  cement,  which 
is  but  little  used.  Hence  the  two  kinds  in  use  in  this  country 
are  Portland  and  Rosendale. 

The  latter  is  a  natural  cement,  and  is  so  called  because, 
while  it  is  now  found  at  a  number  of  places,  it  was  first  found 
at  Rosendale,  Ulster  County,  N.  Y.  It  is  slower  setting  and 
weaker  than  Portland  cement,  and  is  also  cheaper.  Hence 
it  would  naturally  be  used,  on  account  of  its  cheapness,  wher- 
ever great  strength  or  quick  setting  were  not  specially 
required.  On  the  other  hand,  Portland  cement,  which  is  manu- 


724  APPLIED  MECHANICS. 

factured  largely  in  Germany  and  in  France,  and  to  some  extent 
in  the  United  States,  is  made  by  calcining  a  number  of  differ- 
ent mixtures  of  rocks,  but  such  as  will  give  somewhere  in  the 
neighborhood  of  75  per  cent  of  chalk  and  25  per  cent  of  clay. 
Unwin  gives  the  composition  of  Portland  cement  as  lying 
generally  between  the  limits — 

Silica      .     .     ;l7;vil /':';*.     .     20   1026 

Alumina 5    "   10 

Oxide  of  iron       ..<-.-.     .   -v'u  2    "     6 

Lime      .  k*  , ^J-p.Y* v    -.'    /    .     .     67    "  58 
Magnesia    .     .     .     .     <'  >«:  .    Y    .      0.5  "     3 

TESTS  OF  THE  STRENGTH  OF  CEMENTS. 

While  a  good  many  tests  have  been  made  on  the  compres- 
sive  strength  of  cement,  and  a  few  also  on  transverse  or  shear- 
ing strength,  nevertheless  the  test  most  commonly  used  in 
order  to  determine  its  quality  is  the  test  of  its  tensile  strength. 
The  specimen  used  for  this  purpose  is  called  a  briquette,  and 
the  cut  shows  one  of  its  common  forms,  the  smallest 
section  being  generally  one  square  inch.  The  real 
reason  for  using  the  tensile  instead  of  the  compres- 
sive  strength  for  a  test  is,  in  the  opinion  of  the 
author,  that  inasmuch  as  the  tensile  strength  of 
cement  is  very  much  less  than  its  compressive 
strength,  it  follows  that  the  machines  for  testing  the 
tensile  strength  are  cheaper,  and  the  work  of  testing  tensile 
strength  is  less  than  is  the  case  in  testing  its  compressive 
strength,  although  there  are  some  who  give  other  reasons 
which  have  some  appearance  of  plausibility. 

In  order  to  discuss  this  matter  intelligently,  however,  we 
should  bear  in  mind  — 

i°.  Either  the  tensile  or  the  compressive  test  is  compara- 
tive, and  serves  merely  to  determine  the  quality,  and  also 
to  compare  different  lots  of  cement,  but  neither  of  then?  fur- 


HYDRAULIC  CEMENTS  AND  BRICK  PIERS.  ?2$ 

nish  us  any  figures  which  would  be  suitable  to  use  in  computing 
the  allowable  load  on  any  structure  which  depended  upon 
cement  for  its  strength. 

2°.  It  has  been  pretty  well  settled  by  experiment  that  the 
ratio  of  the  compressive  to  the  tensile  strength  of  the  same 
cement  is  about  eight  or  ten. 

From  these  two  facts  we  might  conclude  that,  as  far  as  the 
objects  that  can  be  attained  by  testing  cement  are  concerned, 
either  test  would  answer  the  purpose  equally  well ;  but  inas- 
much as  it  is  also  a  fact  that,  with  the  best  appliances  thus  far 
provided  for  the  purposes,  it  is  possible  to  obtain  greater 
accuracy  in  the  compressive  than  in  the  tensile  test,  therefore 
it  seems  to  the  author  that  the  compressive  and  not  the  tensile 
is  the  test  that  should  be  used  in  making  cement  tests.  Never- 
theless, inasmuch  as  the  tensile  strength  is  most  used,  a  brief 
account  will  be  given  here,  showing  what  has  been  done,  what 
we  can  reasonably  expect  from  good  cements,  and  a  few  pre- 
cautions will  be  mentioned,  which  it  is  necessary  to  use  in 
making  the  tests,  in  order  to  insure  correct  results. 

The  following  list  of  references  to  the  literature  of  cement 
testing  will  be  given  here : 

i°.  Q.  A.  Gillmore:  Practical  Treatise  on  Limes,  Hydraulic  Cements, 
and  Mortars. 

2°.  John  Grant:  Articles  in  the  Proceedings  of  the  British  Institu- 
tion of  Civil  Engineers,  vols.  xxv.,  xxxii.,  and  xli. 

3°  Charles  Colson:  Experiments  on  the  Portland  Cement  used  in 
the  Portsmouth  Dockyard  Extension.  Proc.  Brit.  Inst.  Civ. 
Engrs.,  vol.  xli. 

4°.  Isaac  John  Mann:  The  Testing  of  Portland  Cement.  Proc.  Brit. 
Inst,  Civ.  Engrs.,  vol.  xlvii. 

5°.  VVm.  V.  Maclay:  Notes  and  Experiments  on  the  Use  and  Testing 
of  Portland  Cement.  Trans.  Am.  Soc.  Civ.  Engrs.,  Dec.  1877. 

6°    Eliot  C.  Clarke:  Record  of  Tests  of  Cement  made  for  the  Bos- 


726  APPLIED  MECHANICS. 

ton  Main  Drainage  Works,  1878-1884.  Trans.  Am.  Soc.  Civ. 
Engrs.,  April,  1885. 

7°.  Q.  A.  Gillmore:  Notes  on  the  Compressive  Resistance  of  Free- 
stone, Brick  Piers,  Hydraulic  Cements,  Mortars,  and  Con- 
cretes. 

8°.  J.  Sondericker:  How  to  Test  the  Strength  of  Cements.  Am. 
Soc.  Mech.  Engrs.  for  1888. 

9°.  Bauschinger:  Mittheilungen  aus  dem  Mechanisch-Technischen 
Laboratorium,  Hefte  i.,  vii.,  and  viii. 

10°.  Exec.  Doc.  12,  47th  Congress,  ist  session,  House:  Compres- 
sive Tests  of  Seven  Cubes  of  Concrete. 

11°.  Exec.  Doc.  5,  48th  Congress,  ist  session,  Senate:  Shearing 
Test  of  One  Concrete  Cube. 

12°.  Exec.  Doc.  35,  49th  Congress,  ist  session,  Senate:  Tests  of 
Neat  Cement  and  Cement  Mortars. 

13°.  Preliminary  Report  of  the  Committee  on  a  Uniform  System  for 
Tests  of  Cement.  Trans.  Am.  Soc.  Civ.  Engrs.,  January,  1884. 

14°.  Final  Report  of  the  Committee  on  a  Uniform  System  for  Tests 
of  Cement.  Trans.  Am.  Soc.  Civ.  Engrs.,  January,  1885. 

15°.  Behavior  of  Cement  Mortars  under  various  Contingencies  of 
Use.  F.  Collingwood:  Trans.  Am.  Soc.  Civ.  Engrs.,  Nov. 
1885. 

16°.  Report  of  Progress  by  the  Committee  on  the  Compressive 
Strength  of  Cements,  etc.  Trans.  Am.  Soc.  Civ.  Engrs.,  July, 
1886. 

17°.  Another  Report  of  the  Committee.  Trans.  Am.  Soc,  Civ. 
Engrs.,  June,  1888, 

Of  the  above,  probably  the  one  best  suited  to  give  the 
student  a  fair  idea  of  what  he  may  reasonably  expect  of 
Portland  and  of  Rosendale  cement  is  the  article  ot  Eliot  C. 
Clarke,  inasmuch  as  it  gives  an  account  of  the  results  of  the 
tests  made  to  determine  the  quality  of  the  cement  used  in  the 
construction  of  about  seventeen  miles  of  large  sewers,  a  pump- 
ing-station,  and  a  reservoir,  in  the  building  of  which  there  were 
used  about  180,000  barrels  of  cement. 


HYDRAULIC  CEMENTS  AND  BRICK  PIERS. 

A  brief  summary  of  some  portions  of  the  article  will  be 
given  here. 

The  object  of  the  tests  was  to  determine — 

"  i°.  The  relative  strength  and  value  of  any  cement,  as  com- 
pared with  the  average  strength  and  value  of  the  best  quality 
of  similar  kinds  of  cement." 

"  2°.  The  absolute  and  comparative  strength  and  value  of 
mortars  of  different  kinds  made  from  the  same  cement." 

"  3°.  The  effect  produced  on  cement  mortar  by  different 
conditions  and  methods  of  treatment." 

Clarke  says  that  while  the  absolute  color  indicates  but  little, 
nevertheless  with  Rosendale  cement  a  light  color  generally 
indicates  an  underburned  rock,  and  with  Portland,  underburned 
material  is  indicated  by  a  yellowish  shade,  and  by  a  marked 
difference  between  the  color  of  the  hard-burned,  unground 
particles  retained  by  a  fine  sieve  and  the  finer  cement  which 
passes  through  the  sieve. 

He  gives  the  following  as  weights  per  cubic  foot,  the  pack- 
ing being  accomplished  in  each  case  by  allowing  the  cement  to 
fall  through  a  height  of  three  feet : 

Rosendale      ....'.. 49  to  56  Ibs. 

Lime  of  Teil      , .     .     .  50 

Roman      .     ,».,,..     .     .     .     .     .     .  54 

A  fine  ground  French  Portland    ....  69 

English  and  German  Portlands    .     .     .     .  77.5  to  87    " 

An  American  Portland    .......  95 

"  While  color  and  weight  by  themselves  indicate  little,  yet 
considered  together  and  also  in  connection  with  fineness,  they 
/nabled  the  inspector  to  guess  at  the  character  of  the  cement, 
and*  suggested  reasons  for  high  or  low  breaking.  Thus  a 
cement  which  was  light  in  color  and  weight,  and  also  coarse- 
ground,  would  be  viewed  with  suspicion." 

Clarke  also  calls  especial  attention   to  the  value  of   fine 


APPLIED   MECHANICS. 


grinding,  and  recommends  that  a  No.  120  sieve,  in  which  the 
meshes  are  about  one  two-hundredth  of  an  inch  square,  be 
taken  as  a  standard  ;  and  claims  that  if  we  sift  a  sample  from  a 
barrel  of  cement  by  the  use  of  such  a  sieve,  and  thus  determine 
what  proportion  of  the  barrel  would  pass  through  it,  that  we 
may  fairly  consider  that  the  barrel  contains  that  proportion 
of  cement,  the  rest  being  about  the  same  as  so  much  sand. 

Hence  by  the  use  of  such  a  sieve  we  could  determine — 

1°.  Which  was  really  cheaper  of  two  barrels  which  differed 
only  in  fineness  and  in  price. 

2°.  How  much  sand  must  be  added  in  order  to  obtain  a 
given  proposed  mixture  of  cement  and  sand.  Thus  suppose 
Y9^  of  the  barrel  of  cement  would  pass  the  standard  sieve,  then 
if  we  require  a  mixture  of  one  cement  to  two  sand,  we  should 
add  to  each  barrel  of  cement  only  1.7  barrels  of  sand. 

3°.  Whether  it  is  cheaper  to  pay  for  regrinding  the  cement 
or  to  use  it  with  a  smaller  proportion  of  sand. 

Fine  cement  produced  by  supplying  the  mill  with  compara- 
tively soft  underburned  rock  is  of  course  of  inferior  quality. 

The  strength  of  the  briquettes  after  any  given  length  of 
time  depends,  among  other  things,  upon  the  amount  of  water 
with  which  they  are  gauged.  In  the  (tensile)  tests  made  by 
Clarke,  sufficient  water  was  usually  added  to  make  a  plastic 
mortar,  somewhat  stiffer  than  that  commonly  used  by  masons, 
different  cements  requiring  different  amounts  of  water  to  pro- 
duce this  result.  The  amount  required  approximated  25  per 
cent  for  Portland  and  33  per  cent  for  Rosendale. 

The  following  table,  taken  from  Clarke's  article,  gives  re- 
sults compiled  from  about  25000  breakings,  of  twenty  different 
brands  of  Rosendale  and  of  imported  Portland  cement,  and 
this  table  fairly  represents  the  average  strength  of  ordinary 
good  cements  : 


HYDRAULIC   CEMENTS  AND   BRICK  PIERS.  J2g 


Neat  Cement. 

Cement,  i  ;  Sand,  i. 

Cement,  i  ;  Sand,  1.5. 

8? 

•o 

jj 

! 

0 

a 

VO 

12  mos. 

1 

6 

8 

6  mos. 

12  mos. 

jj 
B 

i 

H 

I 

vo 

| 
N 

71 


ROSENDALE   CEMENT. 


92 


145 


282 


290 


116 


190 


256        41 


95 


155 


230 


PORTLAND  CEMENT. 


102 

303 

412 

468 

494   160 

225 

347 

387 

- 

- 

- 

Cement,  i  ;  Sand,  2. 

Cement,  i  ;  Sand,  3. 

Cement,  i  ;  Sand,  5. 

j»j 

\ 

6 

a 

j 

a 

vo 

j 

x 

V 

U 
£ 

i 

VO 

12  mos. 

jj 

£ 

i  mo. 

6  mos. 

12  mos. 

24 


ROSENDALE  CEMENT. 


60 


125 


180 


35 


80 


121 


16 


46 


80 


PORTLAND  CEMENT. 


126 

163 

279 

323 

95 

140 

198 

257   55 

88 

136 

155 

The  following  are  Clarke's  conclusions  from  this  table : 
"  It  shows  that  Portland  cement  acquires  its  strength  more 
quickly  than  Rosendale  ;  that  both  cements  (but  especially  Ro- 
sendale)  harden  more  and  more  slowly  as  the  proportion  of  sand 
mixed  with  them  is  increased ;  that  whereas  neat  cements  and 
rich  mortars  attain  nearly  their  ultimate  strength  in  six  months 
or  less,  weak  mortars  continue  to  harden  for  a  year  or  more. 
The  table  shows  the  advantage  of  waiting  as  long  as  possible 
before  loading  masonry  structures,  and  the  possibility  of  saving 
cost  by  using  less  cement  when  it  can  have  ample  time  to 


730 


APPLIED   MECHANICS. 


harden.     It  also  shows  that  Portland  cement  is  especially  use- 
ful when  heavy  strains  must  be  withstood  within  a  week." 

The  next  table,  also  from  Clarke,  gives  the  result  of  about 
500  breakings  of  one  single  brand  of  cement,  to  show  the  effect 
of  age,  and  also  of  larger  proportions  of  sand. 


Age  when 

rt    4> 

•'« 

~~   <*• 

M 

H 

--  2 

H 

G     M 

broken. 

v  a 

to  8 

ii 

4)       ~ 

B  ^ 

if 

if 

ii 

!=" 

u 

U  $ 

u  $ 

U  $ 

c!5^ 

d£ 

u  w 

PORTLAND  CEMENT  MORTAR. 

One  week, 

295 

1  66 

89 

50 

33 

23 

17 

One  month, 

34i 

243 

132 

88 

67 

50 

4i 

Six  months, 

374 

343 

213 

149 

98 

76 

5i 

Two  years, 

472 

389 

226 

159 

98 

49 

3i 

ROSENDALE    CEMENT    MORTAR. 

One  week, 

24 

7     i         5 

One  month, 

83 

32 

17 

8 

5 

Six  months, 

172 

93 

62 

50 

33 

21 

Two  years, 

211 

90 

56 

33 

22 

2O 

He  made  some  experiments  to  determine  the  relation  exist- 
ing between  the  stiffness  of  cement  mortar  when  first  mixed 
and  its  subsequent  strength.  "  The  stiffness  depends,  of  course, 
upon  the  amount  of  water. used,  and  varies  somewhat  with  dif- 
ferent cements."  "  The  time  of  setting  is  greatly  retarded  by 
the  addition  of  water."  He  concludes  that  "  the  curves  show 
that  from  20  to  25  per  cent  of  water  gives  the  best  results  with 
Portland  cement,  and  from  30  to  35  per  cent  with  Rosendale ; 
that  the  differences  in  strength  due  to  the  amount  of  water  are 
considerable  at  first,  but  diminish  greatly  with  age  ;  that  the 
soft  mortars,  even  when  semi-fluid  like  grout,  attain  consider- 
able strength  in  time." 

Another  matter  in  regard  to  which  Clarke  made  a  number 
of  experiments,  is  the  fact  that  fine-ground  cements  are  less 


HYDRAULIC  CEMENTS  AND  BRICK  PIERS. 


731 


strong  when  tested  neat,  and  stronger  when  mixed  with  sand, 
than  are  coarse  cements. 

One  other  set  of  tests  made  by  Clarke  will  be  quoted  here, 

as  follows : 

* 

"  Having  occasion  to  build  with  concrete  a  large  monolithic 
structure  in  which  a  flat  wall  would  be  subjected  to  a  transverse 
stress,  it  was  considered  necessary  to  make  experiments  to  find 
the  comparative  resistance  to  such  stress  of  concrete  made 
with  different  cements,  and  with  different  proportions  of  sand 
and  stone. 

"  The  cements  used  in  the  tests  were  an  English  Portland 
and  a  Rosendale,  both  good  of  their  respective  kinds.  Medium 
coarse  pit  sand  was  used,  and  screened  pebbles.  The  beams 
were  10  inches  square,  and  six  feet  long  or  less.  They  were 
made  in  plank  moulds  resting  on  the  bottom  of  a  gravel-pit 
about  four  feet  deep.  After  the  concrete  had  hardened  suffi- 
ciently, the  moulds  were  removed,  and  the  undisturbed  beams 
buried  in  the  pit,  and  left  for  six  months  exposed  to  the 
weather.  They  were  then  dug  out,  and  broken  with  the  results 
•  given  in  the  table.  The  total  breaking-loads  are  given,  includ- 
ing, one-half  the  weight  of  the  beams,  which  averaged  about 
150  pounds  per  cubic  foot. 


Proportion  of  Materials. 

Average  Centre  Breaking- 
weights  in  Pounds. 

Average  Mod- 

ulus Rupture 

Distance  be- 

Distance be- 

in Pounds. 

Cement. 

Sand. 

Stone. 

tween  Supports 

tween  Sup- 

«'4i". 

ports  5'. 

Rosendale,  i 

2 

5 

1782 

690 

67 

i      i 

3 

7 

Beams 

broke  in 

handling. 

Portland,     i 

2 

7 

3926 

J995 

176 

"           i 

4 

9 

3648 

- 

146 

i 

6 

n 

2822 

1190 

112 

732 


A  P PLIED   ME  CHA  A  /CS. 


"  When  transverse  stress  is  to  be  opposed,  it  is  very  impor- 
tant to  give  ample  time  for  the  concrete  to  harden." 

A  few  results  of  compressive  tests  will  next  be  given.  In 
Exec.  Doc.  35,  49th  Congress,  1st  session,  Senate,  is  given  a  set 
of  tests  of  cubes  and  prisms  of  Dyckerhoffs  Portland  cement, 
Newark  Company's  Rosendale  cement,  Norton's  cement,  and 
National  Portland  cement.  The  Dyckerhoff  Portland  cement 
came  from  Amoneburg,  on  the  Rhine,  Germany,  the  age  of  the 
cement  when  tested  being  about  twenty-two  months. 

The  average  resistance  per  square  inch  of  the  different 
specimens  is  given  in  the  following  table : 


CUBES. 


PRISMS. 


No.  of 
Samples. 

Nominal 
Dimensions. 

Ult.  Strength, 
Ibs.  per 
sq.  in. 

6 

i-in.  cube. 

5895 

6 

2    "      " 

7053 

6 

3    "      " 

5892 

6 

4   "      « 

4859 

6 

5    "      " 

4611 

6 

6    "      " 

4272 

6 

7    "      " 

4987 

6 

8    "      " 

4865 

6 

9    «'      « 

4745 

6 

10     "        " 

4761 

6 

n    "      " 

538i 

6 

12     "        " 

5435 

No.  of 
Samples. 

Nominal 
Dimensions. 
Inches. 

Ult.  Strength, 
Ibs.  per 
sq.  in. 

3 

4X4X1 

16460 

3 

4X4X2 

6370 

3 

4X4X3 

6005 

2 

8X8X2 

I  0664  f 

3 

8X8X3 

6168 

3 

8  *8X4 

5952 

3 

8X8X5 

6020 

3 

8X8X6 

5771 

The  following  are  tests  of  mortar  and  concrete  cubes  made 
of  Norton's  cement,  the  numbers  given  for  strength  being  in 
each  case  the  average  of  two  tests.  The  cubes  were  46  months 
old  when  tested. 


HYDRAULIC  CEMENTS  AND   BRICK  PIERS. 


733 


MORTAR. 
Ultimate  strength,  Ibs.  per  sq.  in. 


CONCRETE. 

Ultimate  strength,  Ibs.  per  sq.  in. 


Nominal 

i  Cement; 

i  Cement  ; 

Dimensions. 

1.5  Sand. 

3  Sand. 

4-in.  cube. 

2048 

1325 

6  "   " 

1341 

750 

8  "   " 

1746 

7QO 

12  "    " 

1346 

688 

16  "   " 

1247 

718 

i  Cement; 

i  Cement; 

Nominal 

1.5  Sand; 

3  Sand; 

Dimensions. 

6  Broken 

6  Broken 

Stone. 

Stone. 

4-in.  cube. 

2322 

1635 

6    "      " 

963 

IOOO 

8    "      " 

1433 

862 

12     "        " 

1560 

766 

16    "      " 

1448 

843 

The  tests  of  the  cubes  made  of  Newark  Company's  cement 
will  not  be  quoted,  because  pine  cushions  were  used  for  the 
mortar,  and  also  for  a  part  of  the  concrete. 

The  tests  of  cubes  made  from  National  Portland  cement 
are  given  in  the  following  tables.  These  were  also  46  months 
old,  and  they  are  each  averages  of  two  tests. 


MORTAR. 
Ultimate  strength,  Ibs.  per  sq.  in. 


Nominal 

i  Cement; 

Dimensions. 

3  Sand. 

4  in.  cube. 

3450 

6    "      " 

2155 

8    "      " 

2478 

12     "        " 

2434 

16    "      " 

2520 

CONCRETE. 
Ultimate  strength,  Ibs.  per  sq.  in. 


Nominal 

i  Cement; 

Dimensions. 

3  Sand  • 
6  Broken  Stone. 

4  in.  cube. 

4014 

6    "      " 

2627 

8    "      " 

3027 

12     "        " 

2690 

16   "      " 

2979 

All  the  above-stated  compression  tests  were  made  on  the 
government  testing-machine  at  Watertown  Arsenal. 

Bauschinger  has  also  made  a  large  number  of  compression 
tests,  accounts  of  which  may  be  found  in  Hefte  I,  7,  and  8  of 


734  APPLIED  MECHANICS. 

the  Mittheilungen,  but  for  these  the  student  is  referred  to  the 


Mittheilungen. 


PRECAUTIONS    TO    BE    OBSERVED    IN    TESTING    CEMENTS. 

The  results  obtained  by  testing  different  samples  of  the 
same  -cement  will  vary  with — 

i°.  The  percentage  of  water  used  in  mixing. 

2°.  The  length  of  time  the  sample  has  been  kept  under 
water,  and  also  the  length  of  time  it  has  been  kept  in  the  air 
before  testing. 

3°.  The  temperature  of  the  water  with  which  it  was  mixed, 
and  also  of  that  in  which  it  was  kept ;  also  the  temperature  of 
the  air  in  which  it  was  kept. 

4°.  The  rapidity  of  breaking. 

Hence,  in  order  that  our  results  may  be  of  value,  we  must 
take  pains  to  regulate  all  these  matters. 

But  another  and  all-important  matter  that  has  not  re- 
ceived the  necessary  amount  of  attention  is,  that  some  means 
should  be  adopted  for  distributing  the  pull,  in  the  case  of  a 
tension  test,  evenly  over  the  section  of  the  briquette,  and  in 
the  case  of  a  compression  test  for  distributing  the  thrust  evenly 
over  the  surface  of  the  specimen.  In  the  ordinary  cement-test- 
ing machines  to  be  found  in  the  market  there  is  generally  no 
adequate  provision  for  this  purpose,  and  this  is  the  reason  why 
so  great  a  variation  exists  in  the  results  obtained  with  the 
same  cement,  by  so  many  experimenters.  For  a  fuller  account 
of  this  matter  see  Trans.  Am.  Soc.  Mech.  Engrs.  for  1882,  page 
172. 

BRICKS    AND    BRICK    PIERS. 

The  following  is  a  list  of  references  to  the  tests  of  the 
compressive  strength  of  bricks  and  brick  piers  made  on  the 
government  testing-machine  at  Watertown  Arsenal : 


HYDRA  ULfC  CEMENTS  AND  BRICK  PIERS. 


735 


i°.  Exec.    Doc.  5,  48th   Congress,   ist  session,   Senate,  page   221: 

Bricks  for  Pension  Building. 
2°.  Exec.   Doc.   35,  49th  Congress^   ist  session,   Senate,  page  69: 

Series  of  33  Brick   Piers,  together  with  Bricks  and  Mortar 

used  in  building  them:  page  236,  Six  Brick  Piers. 
3°.   Exec.  Doc.  36,  49th  Congress,    ist  session,   Senate,  page  1138: 

Common,  Hard  Burned,  and  Face  Brick,  such  as  are  used  in 

the  next  series  of  piers. 
4°.  Exec.  Doc.  31,  49th  Congress,  2d  session,  House,  page  1689: 

Another  series  of  53  Brick  Piers. 

It  will  be  sufficient  to  give  here  the  two  series  of  brick 
piers,  together  with  the  tests  of  brick  and  of  mortar  used  in 
their  construction. 

In  building  the  first  series  of  33  piers  three  kinds  of  brick 
were  employed,  viz.,  hard-burned  face-brick,  hard-burned 
common  brick,  and  medium-burned  Bay  State  brick,  all  of 
them  being  made  in  Cambridge,  Mass.  In  testing  all  of  these 
bricks,  the  direction  of  the  pressure  was  at  right  angles  to  the 
largest  face  of  the  brick.  The  following  are  the  results: 


Dimensions. 

Ultimate  Strength. 

Kind 

Compressive 

of 

Area, 

Thick- 

Bricks. 

Length, 

Width, 

Sq.  In. 

Total, 

Per  Sq.  In., 

Inches. 

Inches. 

ness, 

Ibs. 

Ibs. 

Inches. 

( 

7-75 

3-70 

2.10 

28.68 

3I7IOO 

11050 

Face.     .      < 

7-75 

3-70 

2.O7 

29.14 

407500 

13984 

( 

7-52 

3-57 

2.05 

26.85 

449300 

16734 

t 

8.12 

3-50 

2.O7 

28.42 

562300 

19785 

Common.    < 
' 

7.90 

3.56 

2.15 

28.12 

628500 

22351 

8.10 

3-62 

2.12 

29.32 

381000 

12995 

c 

7.90 

3-57 

2.00 

28.20 

313600    ' 

III20 

Bay  State.  < 

7-78 

3-70 

2.02 

28.79 

365900 

12709 

7.70 

3-50 

2.  II 

26.95 

280000 

10390 

736 


APPLIED  MECHANICS. 


"  Three  cubes  of  each  kind  of  mortar  were  made  at  the  time 
and  allowed  to  season  with  the  piers,  and  at  the  age  of  14^ 
months  they  were  tested.  The  age  of  the  plaster  of  Paris 
cube  was  4  months. 

"  When  lime  mortar  is  added  to  cement  in  the  proportions 
here  employed,  its  effect  on  the  mortar  is  found  comparatively 
more  serious  with  the  Portland  cement  than  with  the  Rosen- 


Dimensions. 

Ultimate 

Composition. 

Length, 

Compresse 

i  Surface. 

Strength 
per  Sq.  In., 
T  h<j 

Inches. 

Inches. 

Inches. 

JLiOS. 

Lime,  i;  sand  3  

6.OO 

5-90 

5-80 

135 

Do.            do. 

6.02 

5-82 

5-95 

IIQ 

Do.            do. 

6.00 

5.90 

5.87 

ixfl 

Portland  cement,  i;  sand,  2   . 

6.00 

5-96 

6.00 

560 

Do.                       do. 

6.  ii 

6.00 

5-97 

696 

Do.                       do. 

6.02 

5-95 

6.02 

383 

Rosendale  cement,  I  ;  sand,  2. 

6.04 

5-95 

6.10 

156 

Do.                        do. 

6.08 

6.05 

6.05 

iS6 

Dd.                       do. 

6.22 

6.00 

6.00 

143 

Portland  cement,  neat  .     .     . 

6.  '05 

6.00 

6.02 

2673 

Do.           do. 

6.02 

6.00 

6.03 

3548 

Do.           do. 

6.  10 

5-92 

5-97 

4227 

Rosendale  cement,  neat     .     . 

6.06 

5-98 

6.00 

421 

Do.             do. 

6.  10 

6.00 

6.05 

615 

Do.             do. 

6.  10 

6.14 

6.05 

526 

Portland  cement,  i;  lime  mor- 

tar, 2  (composed  of  lime,  i; 

sand,  ^)                               .  j  '. 

6  o^ 

6.00 

6  .00 

204 

Do.          do.          do. 

W  •  vy  j 

6.00 

6.00 

6.00 

198 

Do.          do.          do. 

6.10 

6.05 

6.14 

175 

Rosendale     cement,    i  ;    lime 

mortar  2             .           . 

6  .00 

e    Q-J 

C      QC 

IQ1 

Do.             do.             do. 

6.00 

0  '  ^Jj 

5-92 

D  •  VD 
5-92 

Ay-t 
193 

Do.             do.             do. 

6.00 

5.92 

5-98 

162 

Plaster  of  Paris    

6.04 

6.'O2 

6.01 

1981 

HYDRAULIC  CEMENTS  AND  BRICK  PIERS.  737 

dale  cement,  weakening  each  to  such  an  extent,  that  practically 
there  is  little  difference  in  the  resulting  mortar,  whether  as 
regards  strength  or  compressibility,  whether  Portland  or  Ros- 
endale  cement  is  used." 

The  table  on  the  opposite  page  gives  the  compressive 
strength  of  the  mortars  used. 

The  compressions  of  these  cubes  under  smaller  loads  than 
the  ultimate  was  also  measured,  but  will  not  be  given  here. 

"  The  cross-section  dimensions  of  the  piers  ranged  from 
8"  X  8"  to  16"  X  1 6",  and  the  heights  from  16  inches  to  10 
feet." 

"  The  bricks  were  laid  on  bed,  and  joints  broken  every 
course,  with  the  exception  of  two  12"  X  12"  piers,  one  of 
which  had  joints  broken  every  sixth  course,  and  one  had  the 
bricks  laid  on  edge. 

"They  were  built  in  May,  1882,  and  seasoned  in  a  cool, 
dry  building.  Their  ages  when  tested  ranged  from  14  to  24 
months." 

"  In  examining  the  results  of  the  cement  and  mortar  tests 
superior  strength  is  found  in  the  Portland  cement ;  likewise  the 
piers  laid  in  Portland  cement  gave  the  highest  resistance,  while 
piers  of  the  same  kind  of  bricks,  but  laid  in  weaker  mortar, 
showed  lower  results,  from  which  it  appears  that  the  quality  of 
the  mortar  employed  is  a  very  important  function  in  the 
strength  of  brick-work."  "  A  material  tested  between  cushions 
weaker  than  itself,  suffers  by  the  disintegration  of  those 
cushions." 


738 


APPLIED  MECHANICS. 


Dimensions. 

Ultimate 

Kind  of 
Bricks. 

Composition  of 
Mortar. 

Height. 

Cross-section. 

Strength 
per 
Sq  In., 

Ft.      In. 

Ibs. 

In. 

In. 

, 

Lime,  i;  sand,  3    .     . 

I     4-13 

7-55 

7-55 

2520 

Do.  i;     do.    3    .     . 

6     8.00 

7.6o 

7.60 

1877 

Portland  cement,  i; 
sand,  2       ... 

I     4.24 

7.60 

7.60 

3776 

Portland  cement,  i; 
sand,  2  .... 

6    8.00 

7.60 

7.60 

2249 

Face.     .- 

Lime,  i;  sand,  3    .     . 
Do.*i;      do.    3  .     . 

i  11.05 
i  11.04 

11.50 
11.50 

11.50 
II.5O 

1940 
1990 

Do.  i;      do.    3  .     . 

9  11.90 

11.50 

11.50 

IS" 

Do.fi;      do.    3  .     . 

9  "-I3 

11.50 

11.50 

1807 

Portland  cement,  i;  i 
sand,  2  .     .     .     . 

i  11.50 

II.5O 

11.50 

3670 

Portland  cement,  i;  1 
sand,  2  .     .     .     .  f 

9  11.30 

11.50 

11.50 

2253 

, 

Lime,  i  ;  sand,  3    . 

i    4.48 

L7-80 

7.80 

2440 

Do.   i;     do.    3    .     . 

6     8.05 

7.90 

7.90 

1540 

Do.  i;      do.    3    .     . 

i  11.90 

11-75 

11-75 

2150 

Do4  1;     do.  3    .     . 

2      0.30 

11-75 

11.90 

2050 

Common  - 

Do.  i;     do.    3    .     . 
Do.§i;     do.    3    .     . 

9    9.60 
10    0.40 

H.75 
11-75 

11-75 
11-75 

1118 
1587 

Portland  cement,  i;  \ 
sand,  2      .     .     .    j" 

10     O.OO 

H.75 

n-75 

2003 

Do.  i;     do.    2   .     . 

2      7-85 

16.00 

16.00 

2720 

- 

Do.  i;      do.    2  .     . 

10      I.  00 

16.00 

16.00 

1887 

' 

Lime,  i;a  sand,  3  .     . 

2      0.10 

12.  IO 

12.10 

1370 

Do.6  i;    do.     3  .     . 

6     0.50 

12.00 

12.00 

H33 

Do.    i      do.     3  .     . 

6     i.oo 

12.00 

12.00 

1210 

Do.    i      do.    3   .     . 

6    0.25 

12.00 

12.00 

1331 

Do.    i      do.    3  .     . 

6     i.oo 

12.50 

12.50 

I2II 

Do.    i      do.    3  .     . 

7  10.25 

12.00 

12.00 

1174 

Bay  State  - 

Do.    i      do.     3  .     . 
Do.    i      do.     3  .     . 

9  n.  60 
10    0.75 

12.00 
8.00 

12.00 
12.  OO 

924 
940 

Do.    i      do.    3  .     . 

10    0.50 

12.  OO 

16.00 

773 

Rosendale,  i;  mortar,  2 

6     1.63 

12.00 

12.00 

1646 

Rosendale,  i;  sand,  2 

6     0.13 

12.00 

12.  OO 

1972 

Portland,  i;  mortar,  2 

6     1.38 

12.00 

12.  OO 

1411 

Portland,  i;  sand,  2  . 

6    0.60 

12.  OO 

12.00 

1792 

Portland  cement,  neat 

6    o.oo 

12.00 

12.00 

2375 

*  Hollow  centre,  4".25  x  4".25. 

t      "  "         4-5x4.5- 

(a)  Joints  broken  every  6  courses. 


+  Hollow  centre,  4.1  x  4.1. 
§        "  "        4.75X4-75. 

(b)  Bricks  laid  on  edge. 


HYDRAULIC  CEMENTS  AND  BRICK  PIERS. 


739 


The  following  table  gives  the  results  obtained  with  the 
bricks  for  the  second  series  of  piers ;  they  are  taken  from  Exec. 
Doc.  36,  49th  Congress,  1st  session : 

COMMON  BRICKS. 


Dimensions. 

Style 
of 
Specimen. 

Color 
of 
Surface. 

Height, 
Inches. 

Compressed 
Surface. 

Modulus  of 
Elasticity 
per  Sq.  In., 
IDS. 

Crushing 
Strength 
per 
Sq.  In., 
Ibs. 

Inches. 

Inches. 

'  ^ 

Dark  red  «     . 

2.08 

7-65 

3.26 

18990 

Single  brick 
on  bed    .  " 

Light  red  .     . 
Medium  red  . 
Do. 

2.18 
2.18 
2.12 

7.84 

7-75 
7.60 

3-37 
3-32 
3-27 

I25OO 
14720 
16530 

Do. 

2.15 

7.64 

3-30 

16350 

Dark  red   .     . 

7.60 

3-35 

2.18 

3750000 

14160 

Light  red  .     . 

7.80    ' 

3-40 

2.28 

2360000 

8970 

Do. 

7.60 

3-36 

2.26 

22OOOOO 

7650 

Do. 

7.90 

3-46 

2.27 

1700000 

7180 

Single  brick 

Do. 

Medium  red  . 

7.68 
7.70 

3-37 
3-33 

2.27 
2.24 

2700000 
2700000 

7880 
10480 

on  end    . 

Do. 

7.58 

3.28 

2.20 

3750000 

8800 

Do. 

7-55 

3-3i 

2.20 

3750000 

12290 

Dark  red  .     . 

7-50 

3-23 

2.20 

3750000 

I3I20 

Do. 

7.48 

3-i8 

2.18 

5000000 

13260 

Do. 

7-52 

3.20 

2.18 

4600000 

15770 

Pile    of     3. 

bricks 

Medium  red  . 
Do. 

6-45 
6.45 

7.60 
7-50 

3-23 
3-25 

.... 

2850 
4290 

Do. 

6.60 

7-53 

3-20 

.... 

4030 

FACE-BRICKS. 


Single  brick  j 
on  bed    .  1 

....  7"" 

2.10 

2.15 

7.70 

7.72 

3-55 
3-57 



13520 
11960 

.... 

2.16 

7.70 

3-57 

.... 

14580 

r 

Dark  red  .     . 

7.70 

3-52 

2.16 

4070000 

8130 

Do. 

7.70 

3-50 

2.14 

3080000 

7970 

Single  brick  1 

Do. 

7-78 

3.60 

2.16 

3150000 

/  ~y  i 

6970 

on  end    .  j 

Do. 

7.68 

3.58 

2.21 

3750000 

7140 

1 

Do. 

7.70 

3-58 

2.10 

3300000 

5700 

I 

Do. 

7.70 

3.62 

2.16 

2500000 

6000 

Pile     of     3J 
bricks  .  .  1 

Do. 
Do. 
Do. 

6.30 
6.60 
6.60 

7.63 
7-83 
7.70 

3-53 
3.60 
3-50 



3830 
3790 
3650 

740 


APPLIED  MECHANICS. 


This  series  comprises  53  piers,  in  the  construction  of  which 
two  kinds  of  brick  were  used,  viz.,  common  hard-burned 
bricks  and  face-bricks,  laid  on  bed,  with  joints  broken  every 
course. 

The  mortar  was  composed  of  Rosendale  cement  i,  sand  2. 
The  piers  were  21  months  old  when  tested.  In  this  series  the 
mortar  was  kept  purposely  the  same  throughout,  so  that  the 
variation  in  strength  should  be  due  to  the  variation  in  dimen- 
sions of  the  piers.  The  mortar,  however,  was  found  to  be 
much  stronger  in  some  places  than  in  others. 

The  quality  of  the  grout  in  the  grouted  piers  was  not  satis- 
factory, being  soft  and  easily  crumbled.  The  following  tables 
give  the  crushing-strength  of  the  mortar  used  in  building  the 
piers : 


Dimensions. 

Ultimate 

Shane 

Strength 

per 

Height. 

Compressed  Surface. 

Sq.  Inch. 

in. 

in. 

in. 

Ibs. 

( 

6.  II 

5-95 

5.90 

157 

Cubes    .     .  -j 

6.!7 

5.98 

5.91 

158 

( 

6.22 

5-90 

5-93 

166 

( 

17-97 

5-95 

5.96 

147 

Prisms  .     .  -< 

17.98 

5-95 

5.96 

153 

( 

18.03 

5-97 

5.96 

149 

The  following  tables  give  the  results  obtained  by  testing  the 
piers  themselves :  — 


HYDRA  ULIC  CEMENTS  AND  BRICK  PIERS, 


741 


FACE-BRICK  PIERS. 


Weight 
per 
Cubic 
Foot. 

Actual  Dimensions. 

Sectional 
Area. 

Ultimate  Strength. 

Height. 

Cross-section, 

Total. 

Per 

Sq.  In. 

Per 
Sq.  Ft. 

Ibs. 

ft.       in. 

in. 

in. 

sq.  in. 

Ibs. 

Ibs. 

tons. 

'132.7 

2     0. 

7.63 

7.6l 

58.06 

141000 

2428 

174.81 

134-4 

2     0.27 

7.64 

7.63 

58.29 

123400 

2117 

152.42 

130.2 

3  n-95 

7-75 

7.68 

59-52 

122016 

2050 

147.60 

129.7 

4     o. 

7-75 

7-70 

59-68 

116000 

1944 

139-97 

127.6 

6     0.37 

7-75 

7-75 

60.06 

117117 

1950 

140.4 

I2Q.6 

6    0.26 

7.85 

7-75 

60.84 

106470 

1750 

126.0 

125.2 

8    0.56 

7.78 

7-75 

60.30 

IO2OOO 

1691 

121.75 

.... 

9  11.27 

7.80 

7-70 

60.06 

100749 

1677 

120.77 

126.8 

10    0.37 

7.82 

7.80 

61.00 

II0500 

1811 

130.39 

126.4 

*5  H.68 

1  1.  60 

ii.  60 

134-56 

257300 

1912 

137.66 

129.0 

5  11-27 

11-55 

11.50 

132.82 

258IOO 

1943 

139.89 

130.3 

5  ii. 

j  11-55 
\    4.20 

11.50) 
4.10  J 

115.61 

219659 

1900 

136.8 

129.5 

5  10.09 

15-45 

15.40 

237-93 

499653 

2IOO 

151.2 

126.0 

5  n.  81 

15-45 

15-35 

237.16 

468700 

1976 

142.27 

COMMON-BRICK  PIERS. 


120.8 

I  10.87 

7-65 

7.52 

57-53 

161000 

2798 

201.45 

123.3 

I  11.13 

7-65 

7.60 

58.14 

157800 

2714 

195.40 

125.4 

4  0.37 

7.60 

7-55 

57-38 

111891 

1950 

140.40 

124.6 

3  11.62 

7.68 

7.58 

58.21 

101867 

1750 

126.00 

121.5 

6  1.62 

7-65 

7.60 

58.14 

144300 

2481 

178.63 

123.3 

6  1.18 

7.60 

7.58 

57-6i 

132503 

2300 

165.60 

121.4 

8  1.50 

7-65 

7.63 

58.37 

90474 

1550 

111.60 

121.4 

7  11.98 

7.60 

7-55 

57.38 

90800 

1582 

113.90 

121.  0 

10  0.93 

7.60 

7.55 

57.38 

86070 

1500 

108.00 

I23.I 

10  1.31 

7.60 

7-55 

57.38 

104200 

1815 

130.68 

123-5 

i  ii.  06 

ii.  60 

11.40 

132.24 

307800 

2327 

167.54 

125.8 

i  10.75 

11.38 

11-35 

129.16 

318500 

2466 

177-55 

124.9 

3  ".58 

n-55 

11.50 

132.83 

224100 

1687 

121.46 

I25.I 

3  11.81 

11.40 

11.30 

128.82 

25"99 

1950 

140.40 

123.2 

6  0.75 

11-45 

11-45 

131-10 

222870 

1700 

122.40 

I2I.7 

6  0.75 

11.48 

11-45 

131.45 

216200 

1644 

118.36 

121.  6 

8  i-37 

n-45' 

11.40 

130.53 

190700 

1461 

105.19 

120.8 

8  0.75 

11.50 

11.40 

131-10 

2IIIOO 

1610 

115-92 

119.5 

10   1.  00 

11-55 

u-45 

132.25 

178200 

1347 

96.98 

*  Core  built  of  common  brick. 


742 


APPLIED  MECHANICS. 


COMMON-BRICK  PIERS — Continued. 


Weight 
per 
Cubic 
Foot. 

Actual  Dimensions. 

Sectional 
Area. 

Ultimate  Strength. 

Height. 

Cross-section. 

Total. 

Per 

Sq.  In. 

Pei 

Sq.  Ft. 

Ibs. 

ft.       in. 

in. 

in. 

sq.  in. 

Ibs. 

Ibs. 

tons. 

126.2 

I    11.00 

j  11.40 

'     4-55 

11.40 
4-50 

109.48 

271500 

2480 

178.56 

127.7 

2      1.  12 

.  U-45 

11.40 

108.23 

265400 

2452 

176.54 

4.90 

4-55 

127.3 

4    0.18 

1   H-35 
4.70 

4.70) 

106.73 

198100 

1856 

133.63 

127.7 

3  11-98 

]  "^80 

11.40) 
4-70) 

107.97 

215200 

1993 

143-49 

II8.8 

6    0.75 

j  H-35 
1    4.90 

n-35  I 
4-60  f 

106.28 

162100 

1525 

110.52 

124.3 

8    3.18 

(  11.50 
"I    4-90 

11.50) 
4.8of 

108.73 

184841 

1700 

122.40 

I25.I 

10    2^.27 

(  11.50 
"j    4.80 

i  i  .  40  ) 

108  .  06 

157500 

1457 

104.90 

123.0 

6     1.18 

15.50 

15.45 

239.48 

358200 

1495 

107.64 

121.  8 

*     6     1.37 

15-70 

15-65 

245-71 

356600 

1451 

104.49 

123.7 

9  ii.  oo 

15.50 

15.40 

238.70 

230200 

964 

69.41 

120.  1 

9  11.98 

15.70 

15.60 

244.92 

247400 

1010 

72.72 

124.3 

6    0.68 

'     8^60 

15-45  I 
8.3of 

167.32 

270100 

1614 

II6.2I 

125.6 

6    0.68 

15.50 
8.70 

15-45  ? 
8.6of 

164.66 

260200 

1580 

II3.76 

123.4 

9  11.25 

15.40 
8.40 

15.40  I 
8.30  f 

167.44 

2I2IOO 

1267 

91.22 

128.9 

9  10.56 

15.45 
8.70 

15-40  I 

8.60  y 

163.11 

202000 

1238 

89.14 

122.3 

*I2      6.5 

1  1.  60 

ii.  60 

134.56 

217200 

1622 

116.78 

125.0 

fl2      6.5 

H.55 

11.40 

131.67 

193300 

1468 

105.69 

II7-I 

{  4    o. 

7.90 

7.90 

62.41 

72300 

1158 

83.37 

124.0 

*3  11.25 

8.00 

8.00 

64.00 

105800 

1654 

119.09 

124.5 

§6     i.i 

j  11-55 

(      4-20 

H.55  j. 

115.76 

60800 

525 

37.80 

*  Laid  with  bond  stones  4  feet  apart. 
t  Common-brick  pier  grouted. 

*  Face-brick  pier  grouted. 

§  Face-brick  pier,  laid  without  mortar. 


.II         00 


FUNDAMENTAL   PRINCIPLES.  743 


CHAPTER   VIII. 
CONTINUOUS  GIRDERS. 

§  244.  Fundamental  Principles.  —  A  continuous  girder  is 
one  that  is  continuous  over  one  or  more  supports  ;  i.e.,  one  that 
has  at  least  one  support  in  addition  to  those  at  the  ends.  The 
principle  of  continuity  is,  that  the  neutral  line  is  throughout  a 
continuous  curve  over  the  supports,  the  tangent  to  one  branch 
of  the  curve  at  the  support  being  a  prolongation  of  the  tangent 
to  the  other  branch. 

Whereas,  in  the  girder  supported  at  the  ends,  the  bending- 
moment  at  the  support  is  zero,  in  the  continuous  girder  there 
is  a  bending-moment  at  the  support,  where  the  girder  is  con- 
tinuous. There  is  also  a  shearing-force  at  each  side  of  the 
support,  the  sum  of  the  shearing-forces  on  the  two  sides  of 
any  one  support  forming  the  supporting-force. 

In  this  chapter  will  be  given  the  general  methods  of  deter- 
mining the  bending-moments,' slopes,  and  deflections  of  con- 
tinuous girders. 

i°.  When  the  loads  are  distributed. 

2°.  When  the  loads  are  all  concentrated. 

3°.  When  there  are  both  distributed  and  concentrated  loads. 

It  is  believed  that  the  reader  will  thus  have  the  means  of 
solving  all  cases  of  continuous  girders,  and  that,  whenever  it 
is  desirable  to  have  a  set  of  simplified  formulae  for  a  small  but 


744  APPLIED   MECHANICS. 

definite  number  of  spans,  or  for  some  special  proportions  or 
distribution  of  the  load,  he  will  be  able  to  deduce  such  simpli- 
fied formulae  from  the  more  general  ones. 

§  245.  Distributed  Loads.  —  In  this  case  we  assume  that 
all  the  loads  are  distributed,  whether  they  are  uniformly  dis- 
tributed or  not.  The  first  step  to  be  taken  is,  to  find  the  bend- 
ing-moment  over  each  support :  this  is  done  by  using  what  is 
known  as  the  "three-moment  equation"  which  we  shall  now 
proceed  to  deduce  ;  and,  in  the  course  of  the  reasoning  by  which 
we  deduce  it,  we  shall  derive  a  number  of  useful  equations,  ex- 
pressing bending-moment,  shearing-force,  slope,  deflection,  etc., 
at  various  points. 

B  Ox  C 


FIG.  247. 

For  the  purpose  in  view,  let  us  assume  our  origin  at  O 
(Fig.  247),  and  let 

Ml  =  bending-moment  at  B. 

M2  =  bending-moment  at  O. 

Mz  =  bending-moment  at  A. 

/x      =  OA. 

/_,    =  OB. 

F0    =  shearing-force  just  to  the  right  of  O. 

F_0  =  shearing-force  just  to  the  left  of  O. 

Ft    =  shearing-force  at  distance  x  to  the  right  of  origin. 

F_i  =  shearing-force  at  distance  x  to  the  left  of  origin. 

Shear  is  taken  as  positive  when  the  tendency  is  to  slide  th& 
part  remote  from  the  origin  downwards. 

If  S0  =  supporting-force  at  (9, 


DISTRIBUTED   LOADS.  745 

Beginning,  now,  by  taking  O  as  origin,  and  x  positive  to  the 
right,  — 

Let  OC  =  x. 

CD  ==  v  =  deflection  at  distance  x  from  origin. 
w     =  load  per  unit  of  length  (either  constant,  or  vari- 
able with  x). 

We  shall  then  have,  from  the  principles  of  the  common 
theory  of  beams, 


F*  =  F0  -J      wdx;  (i) 

i.e.,  the  shearing-force  at  a  distance  x  to  the  right  of  O  is 
found  by  subtracting  from  the  shearing-force  just  to  the  right 
of  O  the  sum  of  the  loads  between  the  section  at  x  and  the 
support ;  and  this  sum  is 


i      wdx. 

«/o  , 


In  a  similar  manner,  if  we  were  to  take  origin  at  O,  and  x 
positive  to  the  left,  we  should  have 


X 


F_Q  —  /     wdx.  '     (2) 


In  §  204  we  found  the  equation 


dM 
—7- 
dx 


dM     w     r*    . 

•    —7-  =  A>  —  I      wdx. 
dx  JQ 


Hence,  integrating  between  x  =  o  and  x  =  x,  and  observing, 
that,  when  x  =  o,  M  =  M2,  we  have 


M  -  M2  =  F&  -  C*  C*  wdx*, 

«/0       t/0 


746  APPLIED   MECHANICS. 

which  reduces  to 

X*  fx 
I      wdx2',  (3) 

t/o 

or,  in  words,  — 

The  bending-moment  at  a  distance  x  to  the  right  of  O  is 
equal  to  the  bending-moment  over  the  support  at  the  origin, 
plus  the  product  of  the  shearing-force  just  to  the  right  of  the 
origin  by  the  distance  of  the  section  from  the  origin,  minus 
the  sum  of  the  moments  of  the  loads  between  the  section  and 
the  support  about  the  section. 

Observe  that  this  sum  of  the  moments  of  the  loads  between 
the  section  and  the  support  about  the  section  has,  for  its  math- 
ematical equivalent,  the  expression 

X 

wdx2 ; 

and,  as  a  particular  instance,  it  may  be  noted,  that  when  the 
load  is  uniformly  distributed,  and  hence  w  is  constant,  this  will 
reduce  to 

WX2  ,         .X 

—  =  («*)  -, 

wx  being  the  load  between  the  section  and  the  support,  and  - 

os  §  nl     2 
being  the  leverage  of  its  resultant. 

Now  write,  for  brevity, 


n 

t/O       t/0 


wdx2  =  m; 

t/o     t/o 

then 

M  =  M2  +  Fjc  —  m.  (4) 

Now,  from  §  194,  we  have 

5?  =  £? 


DISTRIBUTED  LOADS.  747 

Let  ax     =  slope  at  distance  x  to  the  right  of  the  origin. 
a_x  =  slope  at  distance  x  to  the  left  of  the  origin. 
OQ     =  value  of  at  when  .r  =  o. 
a_0  ==  value  of  a_!  when  x  =  o. 
Then 


where  c  is  an  arbitrary  constant,  to  be  determined  from  the 
conditions  of  the  problem. 

If,  now,  we  substitute  for  M  its  value  M2  +  F0x  —  m,  we 
shall  have 

=  d*L  —  ?f  f*  dx 
dx 


:  ^vodr  K  lo^d'"''"1-1"'  j£^ij'r^v;  "     •'•  i 
To  determine  c,  observe,  that,  when  or  =  o,  ox  =  OQ  ; 


x 
dW  tanaI  =  |  =  tana0  +  ^g 


El 


Integrate  again,  and  observe,  that,  when  ;tr  =  o,  v  =  o,  and  we 
obtain 

v  =  ^tanoo  +  M2  r  r^f 

Jo     Jo      •&•* 


748  APPLIED  MECHANICS. 

Now  write,  for  the  sake  of  brevity, 
m  =    I      I    wdx2,     ml  =    (I    o'dk2,      #*_,  =1          I 

«/0      1/0  C/0        «/0  t/o  C/0 


the  last  four  being  derived   by  taking  x  positive  to  the  left. 
We  shall  have 

v  =  ^tanoto  +  M2n  +  F<#  —  V;  (7) 


and,  if  vl  =  deflection  at  A  =  vertical  height  of  A  above  0,  we 
shall  have,  by  substituting  /,  for  x  in  (7), 


Now,  if  we  assume  any  horizontal  datum  line  entirely  below  all 
the  points  of  support,  and  let  the  height  of  B  above  this  line  be 
ybt  that  of  A,  ya,  and  that  of  0,  yoy  etc.,  we  shall  have 


ya  ~  y0  =  /i  tan  oo  +  M2n,  +  F&  -  K,.  (8) 

And,  if  we  put  ^r  =  /,  in  (4),  we  shall  have 

Jf3   =   M2   +  ^o/i   -   ^x 


(9) 


and,  if  we  substitute  this  value  of  F0  in  (8),  we  obtain,  by  redu- 
cing, 


DISTRIBUTED  LOADS.  749 


and,  solving  for  tan  OQ,  we  obtain 


This  expression  gives  us  the  tangent  of  the  slope  at  O  in  span 
OA  ;  and  equation  (9)  gives  us  the  shearing-force  just  to  the  right 
of  O  in  span  OA,  in  terms  of  M2,  M3,  and  known  quantities. 

If  we  were  to  take  the  origin  at  Oy  as  before,  and  x  positive 
to  the  left  instead  of  the  right,  we  should  have,  in  place  of  (4), 

M  =  M2  +  F_0x  —  m;  (n) 

in  place  of  (9), 

f_o  =  •"'-•*>  +  *- 1  (I2) 

and  in  place  of  (10), 


But,  since  the  girder  is  continuous,  we  must  have  the  tangent  at 
O  to  the  left-hand  part,  a  prolongation  of  the  tangent  at  O  to- 
the  right-hand  part,  as  shown  in  Fig.  248. 
Hence  we  must  have 

5  o       T  A 

a_0=     -a.  B 

.*.     tana_0  -f-  tana0  =  o. 

Hence,  adding  (10)  and  (13),  we 

FIG.  248. 

have 

y*-y*     yb  -  y* 


_.       __  __  _  _       _____ 

^-x  «x^  **-,?-.  ^ 

^       "  ""     ""    - 


and  this  is  the  "  three-moment  equation  "  for  the  case  of  a  dis- 
tributed load,  whether  it  be  uniformly  distributed  or  otherwise. 


75°  APPLIED  MECHANICS. 


CASE    WHEN    SUPPORTS    ARE    ON  THE    SAME    LEVEL. 

When  the  supports  are  all  on  the  same  level,  then  y 
yoy  and  the  three-moment  equation  becomes 


/,  I  /_,*    /_<(      /,2       /_,» 
~~  ~ir  ~  7V  +  7t  +  /if =  °- 

MANNER  OF  USING  THE   THREE-MOMENT   EQUATION. 

When  the  dimensions  and  load  of  the  girder  are  known,  all 
the  quantities  in  the  three-moment  equation,  whether  we  use 
(14)  or  (15),  are  known,  except  the  three  bending-moments,  Mu 
M2,  and  My 

Suppose,  now,  the  girder  to  have  any  number  of  (say,  seven) 
points  of  support ;  then,  by  taking  the  origin  at  B  (Fig.  247), 
we  obtain  one  equation  between  the  bending-moments  at  E,  B, 
and  <9,  the  first  of  which,  if  E  is  an  end  support,  is  zero.  Next 
take  the  origin  at  0,  and  we  obtain  one  equation  between  the 
three  bending-moments  at  B,  O,  and  A;  and  so,  continuing,  we 
obtain  five  equations  between  five  unknown  quantities. 

Solving  these,  we  obtain  the  bending-moments  over  the 
supports ;  and  from  these  bending-moments,  after  they  are 
found,  we  can  obtain  the  shearing-forces,  bending-moments, 
slopes,  and  deflections,  by  using  the  equations  deduced  in  the 
course  of  the  reasoning  for  the  three-moment  equation,  as  equa- 
tions (4),  (5),  (7),  (9),  and  (10). 

SPECIAL  CASE, 

when,  the  supports  being  all  on  the  same  level,  the  load  on  any 
one  span  is  uniformly  distributed  over  that  span,  and  when  the 
girder  is  of  uniform  section  throughout. 


DISTRIBUTED  LOADS.  751 

Let  wt    =  load  per  unit  of  length  on  span  OA,  origin  at  O. 

w_i  =  load  per  unit  of  length  on  span  OB,  origin  at  O. 

I     =•  the  constant  moment  of   inertia  of  the  section 

y«  —  yi>  =  y» 
Then 


-  - 


/,3  /_,3 


24^7* 

With  these  substitutions,  the  three-moment  equation,  either 
(14)  or  (15),  becomes 

MJ_t  +  2j/2(/_,  +  /,)  +  MJt  +  J(w«/,3  -h  w.^,3)  =  o.  (16). 

This  is  a  simpler  form  of  the  three-moment  equation,  applicable 
to  this  particular  case  only. 

EXAMPLE   I.  —  Suppose  we   have  a  continuous   girder  of 
uniform  section,  uni- 
formly   loaded,    and    A * *  A 

r  ^i  i  FIG-  249- 

of  three  equal  spans, 

to  find  MB  and  Mc,  also  the  supporting-forces,  shearing-forces, 
bending-moments,  slopes,  and  deflections  throughout. 
Solution.  —  Take  the  origin  at  By  and  we  have 

M,  =  o,  M2  =  M3  =  M*  =  J/c; 

since 

4  =  /.,  =  /, 
equation  (16)  gives 

.-.    M*  =  Mc  =  -— . 
10 


752  APPLIED  MECHANICS. 


Next,  to  find  the  shearing-forces,  we  have,  from  (9), 
wl2       wl2       wl2 

10  10  2        _   Wl 

FB  =  S_C  =  —  -  -, 

equals  shearing-force  just  to  the  right  of  B  or  left  of  C. 
Shearing-force  just  to  the  right  of  C  or  left  of  B  = 

wl2      wl2 
H  --  +  - 

IO  2 


Hence  supporting-forces  are 

SB  =  Sc  =  (f  + 


Bending-moment  in  span  ^4^  at  distance  x  from  ^,  or  in 
span  CD  at  distance  x  from  Z>, 


—  . 


Bending-moment   in   middle   span   at  a  distance  x  from  .Z?  or 
from  C, 

,..  _       wl2       wlx       wx2 


IO  2  2 


Shearing-force  in  span  AB  or  CD  at  a  distance  x  from 
or  D, 

F  =    a//  —  wx. 


Shearing-force  in  middle  span  at  distance  x  from  B  or  C, 

„      wl 
F  =  --  wx. 

2 

Maximum  bending-moment  in  span  BC  (when  x  =  -), 

2 

*,  wl2       wl2       wl2       wl2 

M0  =  ---  j  ----  —  =  -  . 
10          4  8          40 


DISTRIBUTED   LOADS.  753 

Maximum  bending-moment  in  span  AB  or  CD, 
*  =  ¥> 


25  So  25 

Hence   the   greatest   bending-moment  to  which  the  girder  is 

/vpt/2 

subjected  is  that  at  B  or  C,  and  its  amount  is  —  . 

10 

Slope  at  B  in  middle  span,  from  equation  (10), 


tan  ttB  =  __ 


/_   M     ^Y  J_\  _. 


\  _.^L 

i  2E 


10  V     T>El         10  \6JSJ         i  2EI 

__   «//3/_I_    _,J_.  I   \   _         Wl* 

"  ^7\30       60  ~  ^/  ~  120^7' 

which  denotes  an  upward  slope  at  B  towards  the  right.  In  the 
same  way,  the  girder  slopes  upwards  at  C  towards  the  left. 
The  slopes  at  B  and  C  in  the  end  spans  are,  of  course,  down- 
wards. 

Slope  in  the  middle  span  at  a  distance  x  from  B, 

dv          i    (      wlz      .    wlx2       wx*} 
tan  a  =  —  =  -—]  --  x  +  -    ---  —  [  +  c. 
dx       EI\       10  4  6    ) 

When  x  —  o, 

3  * 

tana  = 


W    (     /3  px 

tan  a  =  —  \ 

El  (120       10 


w 


i2oEr 

w 
.*.    Deflection  =  v  s=  WT^ZX  ~~  ^2x*  "+"  I0/^3  ""  51*4) 


754  APPLIED  MECHANICS. 

In  order  to  make  plain  all  methods  of  proceeding,  the  slope 
in  the  end  spans  will  be  found  in  two  different  ways,  as 
follows  :  — 

For  bending-moment,  slope,  and  deflection  in  left-hand  span 
at  a  distance  x  from  B  (or  in  the  right-hand  span  at  distance  x 
from  C),  we  have 

wlz       *  wx2 

M= h  -wlx . 

10        5  2 

dv         i   (      wlzx       $wlx2       wxi\ 

tan  a  =  —  =  — J 4-  * — -  V  4-  c. 

dx       EI\        10  10  6    J 

When  x  =  o, 

wl*  wl* 

~  120EI  "  12QEI 

dv         w  (        /3         /2,#         3  .  #3) 

/.    tan  a  =  —  =  — -\ -  +  ±lx2  -  -  \ 

dx       EI{     120        10        10  6  ) 


l20Er 

/.    Deflection  =  v  =  — ^— -(— A*  —  6l2x2 


We  may,  on  the  other  hand,  accomplish  the  same  object  by 
finding  the  slope  and  deflection  in  left-hand  span  at  distance 
x  from  A,  or  in  right-hand  span  at  distance  x  from  D,  as 
follows  :  — 

dv         i    f  (2    j        wx2  }   ,          w  (lx2  •      #3  )   . 
tan  a  =  —  =  —  I   \  -wlx  --  \dx  =  —  \  ---  >+^. 
dx       EIJ    (5  2    )  EI\s         6) 

When  x  =  l, 


tana= 


I20EI 


,  = 


120^7         30^7  40^7 


DISTRIBUTED   LOADS.  755 


dv          W  (         /3     ,    lx2         x*} 

:.     tan  a  =  —  =  —  \  ---  (  ---  —  I 

//#       .£/(      40         5          6  ) 


The  figure  shows  the  mode  of  bending  of  the  girder. 


FIG.  250. 

To  find  the  greatest  deflection  in  either  span,  put  the  ex- 
pression  for  the  slope  equal  to  zero,  and  find  x  by  the  ordinary 
methods  for  solving  an  equation  of  the  third  degree,  and  then 
substitute  this  value  in  the  expression  for  the  deflection. 

EXAMPLE  II. — Continuous  girder  of  two  equal  spans,  sec- 
tion uniform,  and  load  uniformly  dis- 
tributed. * 
Solution.  —  Take  origin  at  B. 

M,  =  M3  =  MA  =  Mc  =  o,    M2  —  MB,    /,  =  /a  =s  /,    w,  =  w2  =  w; 

therefore,  from  equation  (16), 

<wlz 
— . 

Shearing-force  either  side  of  B  = 

wl2       wl2 

~o~    ~f" 


Supporting-force  at  B  =  \wL 
Supporting-force  at  A  and  C  =•  \wl. 
Shear  at  distance  x  from  A  or  C, 

F  —  \wl  —  wx. 


756  APPLIED   MECHANICS. 

Bending-moment  at  distance  x  from  A  or  C, 

3  wx2 

M=  ±wlx  --- 

8  2 

Maximum  bending-moment  occurs  when  x  =  -|/, 

M0  =  -£*,/>  _  _9_w/»  =  £«£. 
64  128  128 

Hence  greatest  bending-moment  to  which  the  girder  is  sub- 

7£//2 

jected  is  that  at  B,  and  its  magnitude  is  -  . 

8 

Slope  at  B,  from  equation  (6), 

tanaB  =  tana_B  =  -  JL\  _ 

8 


24       12       24 
as  was  to  be  expected. 

Slope  at  distance  x  from  A  in  span  AB, 


When  ^  =  /,      a  =  o  ; 


48^7 

=  ^  =  -^U/**  -  *J  -  £1 
<&       ^7(i6  6        48) 


48^7 
Deflection, 


w     '  •  -  _  8*3  -  fl). 


For  maximum  deflection,  we  have 

^2_-3_Zl 
16  6        48 

/.     A:  =  0.44/. 


DISTRIBUTED   LOADS.  757 

Maximum  deflection  =    w      \  —  i  4-  3  (0.44)  2  —  2(0.44)3}.  (0.44) 
48.fi/  (  ) 


=  -0.0054^. 

EXAMPLE  III.  —  In  order  to  solve  a  case  where  no  simplifi- 
cations enter,  on  account  of  symmetry  or  otherwise,  we  will 
take  a  continuous  girder  of  five  spans  (as  shown  in  the  figure), 
the  spans  varying  in  length  from  3/  to  ///  the  loads  being 
uniformly  distributed,  and  varying  in  intensity  from  $w  on  the 
longest  span  to  *]w  on  the  shortest  ;  the  beam  being  of  uniform 
section. 

A        11        B  4J  C  _  W  _  D  _  «  _  E  _  Tl  _  F 

A         1w        A  6w  A  5w  A  4z0  A  3w  A 

FIG.  252. 

For  this  case  we  can  use  equation  (16). 
Origin  at  B, 

o  +  I4/J/B  +  4/J/c  + 
or 

56J/B  +  i6J/c  =  — 
Origin  at  C, 

4/MB  +  iS/Jfc  +  s/J/p  +  J^[6(64)  +  5(125)]  =  o, 
or 

i6J/B  -f-  72^/0  4-  2oJ/D  =  —iQogwl2. 

Origin  at  D, 

5/J/c  +  22/MD  4-  6/J/E  +  £[5(125)  4  4(216)]^  **  o, 
or 

2oATc  H-  88J/D  4-  24^  =  -^Sgo//2. 

Origin  at  E, 

61M»  +  26/J/E  +  —[4(216)  +  3(343)]  =  o, 

or 

4- 


758  APPLIED  MECHANICS. 

The  four  equations  are  : 

56J/B  +  i6J/c  =  —  573ze//2.  (i) 

I6J/B  +    72^/c  +    20J/D  =  —  I0092£//2.  (2) 

24^  =  —  14892^/2  .  (3) 

-f-  io4J/E  =  —  18932#/2.  (4) 


Eliminate  Ms  between  (3)  and  (4),  and  we  obtain 


Eliminate  MD  between  (2)  and  (5),  and  we  obtain 

2144^  +  8ggSMc  =  —  loionze//2.  (6) 

Eliminate  Mc  between  (i)  and  (6),  and  we  obtain 

234792^  =  -i  769839^/2.  (7) 


/.  from  (i),  Mc  =  —  9.42992^/2, 
from  (5),  M-D  =  —  io.4722ze//2, 
from  (4),  ME  =  —  15.78530//2. 

Shearing-force  just  to  the  right  of 
A  =  ~°  ~ 


-10.4722     +     94299     +     62.5 

— 

9 

-15-7853  +  104722  +  72 


-9-4^99  +  1.53J9  + 

wl  =  12.29152^4 


. 

Wl    = 

+  73-5 


, 
w 


DISTRIBUTED   LOADS.  759 

Shearing-force  to  the  left  of 
7-5379  +  V* 


„       -7-5379  +  94299  +  48 
C  =  -  —  -  wl      —  12.47300/7, 

4 

-9.4299  +  10.4722  +  62.5 
D  —  -    -  —wl  =  12.70840/7, 


8.2450^. 


Supporting-force  at 

A  —     7-9874W/,  C  =  24.  76450/7,  -£  =  25.64050/4 

-5  =  24.5396^,  D  —  23.8229™!,  F  —    8.24500//. 

Shearing-force  at  distance  x  to  the  right  of 

A  in  section  AB  =  7.  98  740/7  —  ywx, 

B  in  section  BC  —  11.52700/7  —  6o/Jt:, 

C  in  section  CD  =  12.29150/7  —  50/3:, 

Z>  in  section  DE  =  11.11450/7  —  ^wx, 

E  in  section  .£7^  =  12.75500/7  —  3o/#. 

Bending-moment  at  distance  x  from 


^4  in  section  ^.5  =  +     7.98740/7*  — 

B  in  section  ^C  =  —   7-5379O/72  -h  11.52700/7^  --   —  , 


C  in  section  CD  =  —  9.42990/72  -f  12.29150/7*;  — 
D  in  section  DE  =  —10.47220/72  +  11.11450/7*  — 
jB  in  section  EF  =  —  15.78530/7*  -j-  12.75500/7*:  — 


760 


APPLIED   MECHANICS. 


For  the  sections  of  maximum  bending-moments  (put  shear- 
ing-force =  o),  — 

In  AB,  x  —  1.  14107,- 
In  BC,  x  —  i.  921  i7; 
In  CD,  x  =  2.45837,- 
In  DE,  x  =  2.77867,- 
In  -£7%  jc  =  4.25177. 

Hence   the   maximum    bending-moments  are  respectively, 
in  — 

Section  AB, 


-9874')= 

+  0/^(11.52707— 


+4.55700/7*. 


-       +  7- 
Section  BC9 

—  7- 
Section 

—  942990/72  +  0/^(12.29157—  fj?)  =    5.678io/72. 
Section  />.£, 

—  io.4722O/72  +  0/^(11.11457  —  2^:)  =    4.9693O/72. 
Section  £"/% 

-I5.78530/72  +  0/^(12.7550   —  J*)  =  11.32970/7*. 

Values  of  tan  a0  —  slope  in  every  case  in  the  span,  towards 
the  right. 

(q,        n\        Mzqt        m,q,         V, 
tana0  =  M2(j~2  --)-    —       -^-  +  -f- 


Slope  at  B, 

o/73/2 
tana.  =  -7-5379^-  -  2    +  9. 


i6o/73 


0/73 


DIS  TRIE  UTED  -  L  OA  DS. 


Slope  at  C, 
tanac    =  - 


625 


Slope  at  D, 
tanaD  =  -10.4722-^(1  -  3)  +  i5-7853^7 

Slope  at  E, 


=  0.7,97^. 


1^7  +  ^177  =  -6.0426 


The  manner  of  bending,  very  much  exaggerated,  is  shown  in 
the  accompanying  figure. 


FIG.  253. 

Slope  at  A  =  -4.096^,     slope  at  F  =  +  23.4578^. 
El  El 

For  the  deduction,  see  what  follows. 
Slopes  in  General. 

Span  AB,  origin  at  A, 

w    (  7 

tana  =  ^-    ' 7' 


7C//3 


When  ^r  =  3/,  tan  a  =  0.3371^—  ; 

W/3  /  x 

••-     ^(35-9433  -  3i-5)  + 


=  —4.106 


El 


w 
/.    tan  a  =  -=-, 


762  APPLIED   MECHANICS. 

Span  BC,  origin  at  B> 

wt3  wl2x    .         ,     wlx2       wx* 

tana  =  0.337^  -  7-5379^  +  5*35^  -  -jj. 


Span-££>,  origin  at  C9 

w  (  c 

tana  =  _  |  -1.5983/3  -  9.4299/2.*  +  6.i4s8/*2  - 

Span  DE,  origin  at  Dy 
tana  =  jL  1  0.7297/3  -  10.4722/2^  +  5.55  725^  -  - 

Span  EFt  origin  at  E, 

tana  =  ^-\  -6.0426/3  -  15.7853^*  +  6.3775^  -   —1 

&1  (  2      ) 

When  ^  =  ;/, 

tana  =  ^—(-6.0426  -  110.4971  +  312.4975  -  172.5) 


Deflections. 
Span  AB, 

w  (                       7 
v  —  ~jjj\  1.3312/^3 x4  —  4.io6A# 

I-1 

Span  BC, 


-  3.7689/2^  H-  i.92ii/#3  _ 

Span  CDj 

w   (  c 

v  =  -grp|  -1.5983/3.1?  -  4.7I49/2**  4-  2.0486/^3  -~* 

Span  Z^.fi', 

»  =  ^|  0.7297/3^  -5.2361/^  +  1.8524  +  ^3  -yl. 
Span  ^7% 

Z;  =  J^J  -6.0426/3^  -    7.8927/2^   +   2.1258/^3   -  —  i 


CONTINUOUS   GIRDER  WITH  CONCENTRATED  LOADS.    763 

The  maximum  deflections  can  be  obtained  by  putting  the 
slopes  equal  to  zero,  as  before. 

§  246.  Continuous  Girder  with  Concentrated  Loads.  — 
For  our  next  general  case,  we  will 

take  that  where  there  are  no  dis-     — ^ 5izl x l- 7r- 

tributed  loads,  but  where  all  the     n-1          <     ><     »          n  +  1 

loads   are   concentrated   at   single 

points,    and   the    section    uniform  Wn~1  Wn 

FIG.  254. 

throughout ;  and  we  will  begin  by 

assuming  only  one  concentrated  load  on  each  span. 

Let  the  support  marked  n  —  i  be  the  (n  —  i)th  support,  and 
the  length  of  the  (n  —  i)th  span  be  /*_,;  let  the  load  on  this 
span  be  Wn_^  and  likewise  for  the  other  spans.  Assume  the 
origin  at  n,  and  let 

F»      =  shearing-force  just  to  the  right  of  n. 
F_n  =  shearing-force  just  to  the  left  of  n. 
Fj      =  shearing-force  at  distance  x  to  the  right  of  n. 
F_  ,  =  shearing-force  at  distance  x  to  the  left  of  n. 
Shear  is  taken  as  positive  when  the  tendency  is  to  slide  the 
part  remote  from  the  origin  downwards. 
If  Sn  =  supporting-force  at  n, 

Sn   =   Fn   +   F_n.  (l)      . 

Let,  also,  xn  =  distance  from  origin  to  point  of  application 
of  load  Wn,  and  let  .*•„__,  =  distance  from  origin  to  point  of 
application  of  load  Wn_^. 

Take  x  positive  to  the  right.     Then,  for 

x<xn,    F,  =  Fn; 

Moreover,  we  have 

dM 


764  APPLIED   MECHANICS. 

hence,  by  integration,  for 

/W  fx 

*<*„,    I     dM  =    I     Fndx; 

jMn  Jo 

(*M  f*x  (*x 

x>xn,   I    dM  =    \    Fndx-    I     Wndx  +  c; 

*)  M  Jo  t/o 


the  value  of  c  being  determined  from  the  condition,  that,  when 
x  =  xn  the  two  results  must  be  identical.     Hence  we  have,  for 


X  <  Xn,        M   —   Mn   ~ 

X>Xn,        M  =   Mn   +  FnX   —    Wn(x   —  X, 

Make  x  =  ln  in  the  last  equation,  and  we  have 
Now  let  4  —  xn  —  #«,  and  (4)  becomes 


hence 


p  =  Mn  + I  ""  Mn 


4 
Moreover,  we  have,  as  before, 


/  being  a  constant. 
Let,  as  before,  — 

04    =  slope  at  distance  x  to  the  right  of  origin. 
o_!  =  slope  at  distance  x  to  the  left  of  origin. 
an    =  value  of  a,  when  .r  =  o. 
a_w  =:  value  of  a_x  when  x  =  o. 


CONTINUOUS   GIRDER  WITH  CONCENTRATED   LOADS.   76$ 

Then  by  integration,  determining  the  constant  in  the  same 
way  as  in  (3),  we  have,  for 

x2 

x  <  xn,     ^"/(tan  a,  —  tan  a«)  =  Mnx  +  Fn—  j 

-r2 

x  >  xn,     £/(tan  a,  -  tan  a«)  =  Mnx  +  ^- 

2 

Hence 

*  <  xn,    EId-j-  =  Elta&on  +  ^«*  +  ^«-; 

d&C  2 

*  >  *„     £/       =  El  tan  a. 


Integrate  again,  and  determine  constants  in  the  same  way, 
and  for 


(8) 


x  <  xn,      EIv  = 

r?7-  z?7-    4. 

x  >  xn,      EIv  =  EIx  tan  aw 


Make  x  =  ln  in  the'  last  equation,  and  denote  the  heights 
of  the  supports  above  the  datum  line  in  the  same  way  as  in 
§  245,  and  we  have 

EI(yn  +  i  -  yn)  =  EIIK  tan  an 


Substitute  for  4  —  -^«>  ^«»  and  for  Fn)  its  value  from  (6),  and  we 

have 

EI(yn  +  ,  —  yn)  =  EIln  tan  an 

/  2  /  2  H^/7 

+  #£-  +  Mn  +  ^  +  ^(b  -  ^2)«  (10) 

3  66 

Hence 


6 

=  El(yn+l  ~y"\     (n) 
4 


766  APPLIED   MECHANICS. 

Now,  if  we  take  origin  at  n  and  x  positive  to  the  left,  we 
should  obtain,  instead  of  (n), 

El  tan  a_«  + 


Now  add  (11)  and  (12),  and  observe,  that,  since  the  girder  is 
continuous, 

tan  a^  -f  tan  a_n  =  o, 

and  we  obtain 

—(4  +  4-,)  +  Mn_^  +  ^%^ 
3  66 


;        (I3) 


and  this  is  the  "three-moment  equation"  for  the  case   of   a 
single  concentrated  load  on  each  span,  and  a  uniform  section. 
When  the  supports  are  all  on  the  same  level,  this  becomes 


^(4  +  4.x)  +  M^fte*  +  -^±^  +  -^p(42  -  as) 
3  6  6  64 

+  ^"/^~I(4-I2  -  an_*)  =  o.   (14) 

64  -i 

Either  of  these  equations  can  be  used  (when  it  is  appli- 
cable) just  as  the  three-moment  equation  was  used  in  the  case 
of  distributed  loads. 


CASE    OF  MORE    THAN  ONE  LOAD    ON  EACH  SPAN.     767 


CASE   OF    MORE   THAN   ONE   LOAD   ON    EACH   SPAN. 

When  there  is  more  than  one  load  on  each  span,  the  three- 
moment  equation  becomes  as  follows  :  — 


+    ^        n&n  /  /  2  x»  2\      i      ^        n  —  I    n  —  r  /  /         2  a\ 

S-7T-  (42  -  ««2)  +  2  —  —  --  (4  -i2  -  an-i2) 
64  64  _  i 


4  - 1 


In  using  these  equations  for  concentrated  loads,  we  can 
determine  the  moments  over  the  supports  ;  but  we  must  observe, 
that,  in  getting  slopes  and  deflections,  bending-moments,  etc., 
the  algebraic  expressions  that  represent  them  are  different  on 
the  two  sides  of  any  one  load,  and  hence  we  must  deduce  new 
values  whenever  we  pass  a  load,  determining  the  constants  for 
our  integration  to  correspond. 

EXAMPLE.  —  Given  a  continuous  girder  of  three  spans,  the 
middle  span  =  20  feet,  each  end  span  =  15  feet;  supports  on 
same  level.  The  only  loads  on  the  girder  are  two ;  viz.,  a  load 
of  5000  Ibs.  at  5  feet  from  the  left-hand  end,  and  one  of  4000 
Ibs.  5  feet  from  the  right-hand  end.  The  supports  are  lettered 
from  left  to  right,  A,  B,  C,  D,  respectively.  Find  the  greatest 
bending-moment  and  greatest  deflection. 

Solution.  —  Origin  at  B, 


Origin  at  C, 
#c/ 

-(20  + 


768  APPLIED   MECHANICS. 


These  reduce  to 

1000000 


nf       , 

4-  2oMc  -\  --      —  =  o 


800000 


o 


MB  =    —4000  foot-lbs. 


Mc  =  —2667  foot-lbs. 


3 

Shearing-forces.  Supporting-forces.  Slopes  at  supports. 

3067,     F_c  =  —67.         SA  =  3067.          tan  aA    =  — 


yV_B  =  1933,    FC    =1511.         oB  =  2000.          tan  aB     =  4- 
Fz     =      67,     F_v  =  2489.         Sc  =  1444.          tan  ac     =  — 

SD  =  2489.          tan  a_D  =  — 

y^y 

Span  AB>  origin  at  A, 

x  <  5,    M  =  30673:. 

x>  5,     M=  30673:  -  5000(3:  -  5)  =  25000  -  19333:. 

Maximum  bending-moment  occurs  when  x  ==  5  and  therefore 
MQ  =  15333. 

x  <  5,     ^7 tan  a,  =  -59444  +  I533-*2; 

x  >  5,     y5"7tan  ax  =  250003:  —  9673^  4-  c. 

Determine  c  by  condition,  that,  when  x  =  5,  these  two  become 
equal ; 

/.     c  =  -121944; 

.*.     x  >  5,     y5"7tan  at  =  —121944  +  250003:  —  9673:2. 

For  deflections, 

x  <  5,     2?7z;  =  —   594443:  4-  5H3:3; 

3:  >  5,     yY7z;  =  —1219443:  -f  125003:2  —  3223:2  -f-  c. 

Determine  c  from  condition,  that,  when  x  =  15,  v  =  o; 

.-.     <r  =  104167; 

/.    3:  >  5,     y5"/y  =  104167  —  1219443:  4-  125003^  —  3223:2. 


CASE   OF  MORE    THAN  ONE  LOAD   ON  EACH  SPAN.    769 

For  maximum  deflection,  equate  slope  to  zero,  and  find  x. 
We  find  it  at  x  —  6.53. 
.-.    EIv0  =  —249531. 

Span  BC,  origin  at  B, 

M=  —4000  +  67*, 
El  tan  a,  =  35557     —  4000*  +  33*% 
EIv          =  35557*  —  2OOO*2  -f-  n.#3. 

For  maximum  deflection,  equate  slope  to  zero,  and  find  x. 

We  find  it  at  x  =  9.78. 

/.    EIvQ  =  166740. 
Span  CD,  origin  at  C, 
x  <  10,  M  =  —2667  +  1511*,' 

*>  10,  M  —  —2667  +  1511^  —  40o°(*  —  10)  =  37333  —  2489,*; 
x  <  10,  jEJtan  ax  =  —31111  —  2667^  -f  756^; 
x>  10,  El  tan  a,  =  —231111  +  37333*  —  1245**. 

For  deflections, 

x  <  10,  EIv  =  —  31111*  —     I334*2  -h  252*2  • 

x  >  10,  EIv  =  —231111*  -|-  18667*2  ~~  4I5-^3  +  £• 

When  x  =  15,  v  =  o; 

.-.    JC>IQ,    EIv  =  —37132785  —  231  in*  +I8667*2  —  415*3. 
For  maximum  deflection,  equate  slope  to  zero,  and  find  x. 

We  find  it  at  *  =  8.41. 
/.     EIv0  =  —24506. 

Hence  greatest  bending-moment  and  greatest  deflection  are 
both  in  span  AB. 

Observe,  that,  since  we  have  used  one  foot  as  our  unit  of 
measure,  all  dimensions  must  be  taken  in  feet,  and  the  value 
of  E  is  also  144  times  that  ordinarily  given. 


770  APPLIED  MECHANICS. 

§  247.  Continuous  Girder,  with  both  Distributed  and 
Concentrated  Loads.  —  In  this  case  we  may  either  calculate 
the  bending-moments,  slopes,  and  deflections  due  to  each  sepa- 
rately, and  then  add  the  results  with  their  proper  signs,  or  we 
may  modify  the  solution  that  was  used  for  the  case  of  a  dis- 
tributed load,  so  as  to  extend  its  applicability  to  this  case. 

Let  W  represent  any  one  concentrated  load,  and  let  x^  rep- 
resent the  distance  of  its  point  of  application  from  the  origin. 
Then,  in  the  general  formulae  deduced  for  the  distributed  load, 
make  the  following  changes  ;  viz.,  — 

i°.  Instead  of 


=/T 

t/o    t/o 


put 

t/o    t/o 

since,  as  was  shown,  m  represents  the  sum  of  the  moments  of 
the  loads,  between  the  section  and  the  support,  about  the  sec- 
tion. 

2°.  Instead  of 


put 


and  make  the  corresponding  changes  in  the  values  of  miy  m_t, 
Viy  and  F_r,  leaving  n  and  q  just  as  before  ;  then  use  the  same 
three-moment  equation  as  before,  with  these  substitutions,  i.e., 


SPECIAL    CASE. 


SPECIAL    CASE, 

when  the  distributed  load  is  uniformly  distributed  on  each  span, 
but  may  be  different  on  the  different  spans,  and  when  the  girder 
is  of  uniform  section. 

Let  w1    =  weight  per  unit  of  length  on  OA. 
w_I  —  weight  per  unit  of  length  on  OB. 

Denote  by  WI  any  concentrated  load  on  OA  at  distance  XT 
from  O. 

Denote  by   W_I  any  concentrated  load  on  OB  at  distance 
<#_!  from  O. 

Then  we  shall  have 


w  J  t2 


v      _  w__ 

~ 
and,  as  before, 


Making  these  substitutions  in  the  three-moment  equation, 
and  clearing  fractions,  we  obtain  for  the  case,  when  the  sup- 
ports  are  all  on  the  same  level, 


o  =  _ 

--  (4  - 


772  APPLIED  MECHANICS. 

CONCENTRATED    AND    DISTRIBUTED    LOADS. 

EXAMPLE.  —  Let  the  girder  be  of  uniform  section,  of  two 
equal  spans,  each  being  10  feet ;  let  the  concentrated  loads  be 
A  B  c  as  shown  in  the 

4'.   'l<  >       >U    4'     >[<    3-  >  figure,    the     dis- 
*  tributed  load  be- 

1000  100°  .  ,T   ,, 

mg    96    Ibs.  per 

FlG-255'  foot.      Find    the 

value  of  El,  so  that  the  deflection  may  nowhere  exceed  -^-§  of 
the  span. 

Solution.  —  Use  equation  (12);  and,  in  deducing  value  of 
J/B,  use  dimensions  in  feet ;  afterwards  use  inches. 

Origin  at  B,  **•        , -. 

MA  =  Mc  =  o; 

-j-  -9?6-(iooo  +  1000) 

u  1000(51) (7)  +  1000(91) (3)f  =  o, 


>U 


40^/3  +  48000  -f-  139800  =  o,     or    MB  =  —4695  foot-lbs., 

MB  =  —56340  inch-lbs.,    MA  =  Mc  =  o. 
mt  =  177600,  m_T  =  201600, 

7200  nl        60  7200  «_T       60 

"*   ''  :  ~EI'  7t=£l'  n~*  ="£7'  T^E~I 

288000  ^         20  288000          q_i        20 

^    :     ~ET          ^  =  ^7'  q~*  =    ~Ef~         U  =  EI' 

175680000   F!  ^  1464000  193536000  V_i    I6I2800 

~~  '      =   ~~  -1  =    ~~  '       =  ~~ 


Shear  right  side  of  middle  =  °  +  =  1949-5, 

Shear  left  side  of  middle      =  °  +  g634°  +  2°l6o°  =  2149.5  ; 

120  ^y  °  ' 


Shear  left  end  =  ~5634°  +  I536°° 


I2O 


810.5, 


c,i_  •    t_  j  —56340   +    177600 

Shear  right  end  =  — 2-^Z LL _.  1010.5  ; 

Middle  supporting-force       =  4099. 


SLOPES.  773 


B ending-Moments  in  Each  Span. 
Span  AB,  origin  at  A, 

810.5.*:  —  43? 
or 

810.53;  —  43?  —  2ooo(#  —  72). 

Span  BC,  origin  at  Z?, 

-56340  +  I949-5*  -  4*2, 

—  56340  +  1949-5.*  —  4*2  ~  1000 (j?  —  36), 

—  56340  4-  1949.5^  —  4x*  —  iooo(^  —  36)  —  iooo(#  —  84). 


To  ascertain  position  of  the  greatest  bending-moment,  dif- 
ferentiate each  one. 

810.5  —  8x  —  2000                =o,  x  =  a  minus  quantity; 

1949.5  -8*  =  o,  x  =  243.69; 

1949.5  —  8*  —  1000                 =o,  x  ==  118.69; 

1949.5  —  8x  —  1000  —  1000  =  o,  x  =  a  minus  quantity. 

Hence,  in  span  AB,  maximum  bending  is  at  the  load,  and 
its  amount  is 

(810.5)  (72)  -  4(72)  (72)  =  37620. 

Span  BC,  maximum  is  at  right-hand  load,  and  is 
-56340  +  1949-5(84)  -  4(84)  (84)  -  1000(48)  =  31194. 

SLOPES. 

Slope  at  B, 

^_  -56340 ,  .       (177600)  (20)   t   1464000      165600 

El     ^20  ""  to>  ~ 

Slope  and  Deflection  in  Span  AB. 
First  part, 

I    (  4      ) 

tan  a  =   -=r  \  405.25^  —  -  #3  I 

«o  being  slope  at  A. 


7/4  APPLIED  MECHANICS. 

Second  part, 
tana  =  -gy  j  405.25**  —  -*s  —  looo*2  +  144000*!  +  c. 

When  x  =  72,  a  is  the  same  in  both  cases  ; 

/.    JLjiooo(72)2  -  (144000X72)!  +  tanoo  -  c  =  o 

<tLl 

5184000 

A    c  =  tanoo  --  -£/  — 
When  x  =  120,  the  second  value  of  tana  becomes      t.r 


5184000        165600 

tan«.-    --  =  -_ 


tan  «<„  =  -^{  -6411600  +  5184000  +  i6S6ooj  =  - 

t  p,,  ^T~~-X/?-   .  >,r     ..;..,.,  '.^  r,T    ' 

•         _        6246000 
~~ET 

Hence  slope  in  first  part  (between  A  and  the  load), 

i    (  4     ) 

tana  =  -pj\  —1062000  +  405.25^  --  #3>« 

Second  part  (between  B  and  the  load), 

i    (  4 

tana  =  ~^r\  —6246000  -f  144000.*  —  594-7S*2  --  -#3 

Deflection. 
First  part, 

v  —  -^-  J  —  1062000*  -f  135.08*3  —  £L(. 
^y  (  3  ) 

Second  part, 

»  =  ~-  1  —6246000*  -t-  72000**  —  198.25*5  —  -*n  - 
^/(  3     ) 


SLOPES.  775 


When  x  =  1 20,  v  =  o ; 

c  = 

(l2O)4) 
(6246000)  ( 1 20)  -  (72000)  (I20)2+  (198.25)  (l20)3  +  ^-y^ 

' 


Point  of  greatest  deflection  is  found  by  putting  slope  equal 
zero.  Moreover,  it  is  plain  that  the  greatest  deflection  is  in  the 
first,  and  not  the  second,  part. 

Hence  equation  is 

^J*3  —  405. 25**  -f-  1062000  =  o 
.'.    *=S6".77J 

and,  substituting  this  in  the  expression  for  the  deflection,  we 
obtain 

39037720 
El 

i 20       39037720 

Hence,  putting  —  = ~j — ,  we  obtain 

/J.OO  xiy 

£1=  130125733. 

If  E  =  1400000,  I  =  92.9 ;  therefore,  if  b  =  3  inches,  h  = 
7  inches. 

Slope  and  Deflection  in  Span  BC. 
Portion  nearest  B, 

tana  =  -gj\  165600  —  56340*  +  974.8** ; 

When  x  =  36  inches,  we  obtain 

i                                                                                661507 
tana  =  -^y  (165600  —  2028240  +  1263341  —  62208)  = pj    • 

Middle  portion, 

1    (  4     ) 

tana  =  -=j\  -20340*  +  474-7S*2  -  -*3}  +  '• 


7/6  APPLIED  MECHANICS. 

661507 

When  x  =  36  inches,  then  tan  a  =  —  -pj  —  ; 

/.     —661507  =  —732240  +  615276  —  62208  H-  EIc 

482335 


.:    tana  =  -^{-482335  -  20340*  +  474-7S*2  -  ^J- 
When  x  =  84  inches, 
tana=  -^(-482335  -  1708560  +  3349836  —  790272) 

368669 
~~ET 
Portion  nearest  C, 

i  =  63660*  -  25*'  -  -*3     +  c. 


When  x  =  84  inches,  then  tan  a  =   ~,  J 

.*.  368669  =  534744°  —  176400  —  790272  4- 

4012099 


/.    tana  =  ^    —4012099  +  63660^ 

When  x  =  120  inches, 

i  963101 

tana  =  -gj(  -401  2099  +  7639200  —  360000  —  2304000)  =  —  -gj 

Deflection. 

Portion  nearest  B, 

V  =  -^-{  165600*  —  28I7O*2  -f  324.9*3  —  1*4  I. 

EI  (  3  ) 

When  x  =  36  inches, 

v  =  —^(165600  -  1014120  +  421070  -  15552)  (36) 

(443002)36  _   15948072 
El         El 


SLOPES. 


777 


Middle  portion, 

2335*  -  10170 

594     72 


v  =  ^  |  -482335*  -  10170*2  +  158.25*3  -  -*4  -f  * 


When  x  =  36  inches,  then  v= 

•'•     -15948072  =  (-482335  -  366120  +  205092  -  15552)  +  EIc> 

15289157 


••'•  v  =  -~j|  -i5289r57  -  482335*  —  IOI70*2  +  158.25*3  -  l*4J. 
^  3      * 

Greatest  deflection  occurs  in  the  middle  portion,  and  the 
point  is  given  by  the  equation. 

o  =  -482335  -  20340*  +  474-7S*2  —  i*3  =  o; 

/.    *  =  71.4. 
Greatest  deflection  in  span  BC, 

FIG.  256. 

V  = 

•^(-15289157-34438719-51846253+57602105-8662899) 

52634923 


I  20       52634023 
Hence,  putting  —  =  -    %y  °,  we  obtain 

4OO  £Ll 

EI=  175449743; 

therefore,  if  E  =  I4CXDOOO,  we  have 

/=  125.3. 
If  b  =  3  inches,  h  =  8  inches. 


El 


778  APPLIED  MECHANICS. 


EXAMPLES   OF   CONTINUOUS   GIRDERS. 

i°.  Let  /  =  uniform  moment  of  inertia  of  girder. 

w  =  load  per  unit  of  length  uniformly  distributed. 
Find  expressions  for 

1,  the  bending-moment  over  each  support, 

2,  the  supporting-forces, 

3,  the  greatest  bending-moment, 

4,  the  slopes  at  the  supports, 

5,  the  greatest  deflection, 

in  each  of  the  following  cases  :  — 

(a)  Two  equal  spans,  length  /. 

(b)  Three  equal  spans,  length  /. 

(c)  Four  equal  spans,  length  /. 

(d)  Two  spans,  lengths  /,  and  12  respectively. 

(e)  Three  spans,  lengths  /„  /2,  and  /3  respectively. 
(/)  Four  spans,  lengths  /„  /2,  /3,  and  /4  respectively. 

(g)  Two  equal  spans ;  loads  per  unit  of  length  on  each  span, 
wl  and  w2  respectively. 

(h)  Three  equal  spans  ;  loads  per  unit  of  length  on  each  span, 
w»  w2>  and  wz  respectively. 

2°.  Do  the  same  in  the  case  where  each  span  is  loaded  with 
a  centre  load  W,  and  has  no  distributed  load. 

3°.  Find  greatest  bending-moment  and  greatest  deflection 
for  a  continuous  girder  of  two  spans,  uniformly  loaded  on  these 
two  spans  with  load  w  per  unit  of  length,  and  which  overhang 
the  outer  supports ;  the  overhanging  parts  having  lengths  /_0 
and  /0  respectively,  and  the  same  distributed  load  per  unit  of 
length  on  the  overhanging  parts. 


EQUILIBRIUM  CURVES.  —  ARCHES  AND   DOMES.         7/9 


CHAPTER    IX. 
EQUILIBRIUM  CURVES.— ARCHES  AND  DOMES. 

§  248.  Loaded  Chain  or  Cord. —  It  has  been  already  shown 
(§  126),  when  the  form  of  a  polygonal  frame  is  given,  that  the 
loads  must  be  adapted,  in  direction  and  magnitude,  to  that  form, 
or  else  the  frame  will  not  be  stable.  The  same  is  true  of  a 
loaded  chain  or  cord,  which  would  be  realized  if  the  frame  were 
inverted. 

If  a  set  of  loads  be  applied  which  are  not  consistent  with 
the  equilibrium  of  the  frame  under  that  form,  it  will  change  its 
shape  until  it  assumes  a  form  which  is  in  equilibrium  under  the 
applied  loads. 

As  to  the  manner  of  rinding  (when  a  sufficient  number  of 
conditions  are  given)  the  stresses  («) 

in  the  different  members,  etc.,  this 
was  sufficiently  explained  under  the 
head  of  "  Frames,"  and  will  not  be 
repeated  here,  as  the  figures  speak 
for  themselves. 

In  Fig.  257  the  polygon  fedcbaf 
is  the  force  polygon,  while  the  equilibrium  polygon  is  123456, 
an  open  polygon.  A  straight  line  joining  e  and  a  would  repre- 
sent the  resultant  of  the  loads. 


FIG.  257. 


78o 


APPLIED   MECHANICS. 


CHAIN   WITH   VERTICAL  LOADS. 

If  all  the  loads  are  vertical,  the  broken  line  edcba  becomes 

a  straight  line  and  vertical,  as 
shown  in  Fig.  258^.  Wheneverthe 
loads  are  concentrated  at  single 
points,  as  2,  3,  4,  5,  the  form  of 
the  chain  is  polygonal ;  and  when 
the  load  is  distributed,  it  becomes 
a  curve,  as  shown  in  Fig.  259. 


FIG.  258. 


CURVED  CHAIN  WITH  A  VERTICAL  DISTRIBUTED  LOAD. 

Given  the  form  of  the  chain  AOE  supported  at  A  and 
and  the  total  load 
upon  it  (be,  Fig. 
259^),  to  find  the 
distribution  of  the 
load  graphically. 
First  lay  off  be  to 
scale,  to  represent 
the  total  load  :  this 
is  balanced  by  the  two  supporting-forces  at  A  and  E  respec- 
tively, as  shown  in  the  figure.  Hence  draw  ca  parallel  to  the 
tangent  at  E,  and  ba  parallel  to  that  at  A,  and  we  have  the 
force  polygon  abca;  the  equilibrium  curve  being  the  chain  AOE 
itself.  Moreover,  if  the  lowest  point  of  the  chain  be  O,  then 
the  load  must  be  so  distributed  that  the  portion  between  O  and 
A  shall  be  balanced  by  the  tension  at  O  and  that  at  A,  and 
hence  that  its  resultant  shall  pass  through  the  intersection  of 
the  tangents  at  O  and  A.  Its  amount  will  be  found  by  drawing 
from  a  a  horizontal  line ;  and  then  we  shall  have  ao  as  the  ten- 
sion at  oy  ab  as  the  tension  at  A,  and  bo  as  the  load  between  A 
and  O.  Hence  the  load  between  E  and  O  will  be  oc. 


LOADED   CHAIN  OR   CORD.  781 

Moreover,  the  load  between  O  and  any  point,  as  B>  will  be 
balanced  by  the  tension  at  O,  and  the  tension  at  B>  and  hence 
will  be  od,  where  ad  is  drawn  parallel  to  the  tangent  BD,  so 
that  the  load  between  B  and  E  will  be  dc ;  and  in  this  way  we 
see  that  we  can  find  the  tension  at  any  point  of  the  chain  by 
simply  drawing  a  line  from  a,  parallel  to  the  tangent  at  that 
point,  till  it  meets  the  load-line  be. 

It  is  to  be  observed,  that,  if  the  tension  at  any  point  of  the 
chain  be  resolved  into  horizontal  and  vertical  components,  the 
horizontal  component  will,  when  the  loads  are  all  vertical,  be  a 
constant,  and  the  vertical  component  will  be  equal  to  the  por- 
tion of  the  load  between  the  lowest  point  and  the  point  in 
question. 

If  we  assume  our  origin  at  <9,  axis  of  x  horizontal  and  axis 
of  y  vertical,  and  let  the  co-ordinates  of  B  be  x  and  y,  and  if  w 
be  the  intensity  of  the  load  at  the  point  (#,  j/),  we  shall  have,  for 
the  load  od  between  O  and  B> 

P  =  fwdx; 

O 

and,  since  the  angle  oad  =  angle  BDC,  we  shall  have 

dy       BC  =  od       P 
dx  ~~  DC  ~~  ~oa  ~~  ~H 

By  differentiation,  we  shall  have 


or 

d*         w 


(T\ 
(0 

and  this  is  the  equation  for  all  vertically  loaded  cords. 


782 


APPLIED   MECHANICS. 


From  it  we  can  find  the  form  of  the  cord  to  suit  a  given 
distribution  of  the  load. 

§  249.  Chain  with  the  Load  Uni- 
formly Distributed  Horizontally.  — 
In  this  case  w  is  a  constant  ;  and  if 
we  assume  our  origin  at  the  lowest 
point  of  the  chain,  and  use  the  same 
notation  as  before,  we  shall  have 

d2y  _  w 


Hence,  integrating,  and  observing,  that,  when  x  =  o,  —  =  o, 

we  have 

dy  _  wx  . 

~          ~ 


and  by  another  integration,  observing,  that,  when  x  =  o,  y  =  o, 
we  obtain 


This  is  the  equation  of  a  parabola;  hence  a  chain  so  loaded 
assumes  a  parabolic  form. 

EXAMPLE  I.  —  Given  the  heights  of  the  piers  for  support- 
ing a  chain  so  loaded,  above  the  lowest  point  of  the  chain,  as 
8  and  18  feet  respectively,  the  span  being  100  feet,  to  find  the 
distance  of  the  lowest  point  from  the  foot  of  each  pier,  and 
the  equation  of  the  curve  assumed  by  the  chain. 

Solution.  —  If  (with  the  lowest  point  of  the  chain  as  origin) 
we  call  (x»y?)  the  co-ordinates  of  the  top  of  the  first  pier,  and 
(*2>  ?2)  those  of  the  top  of  the  second  pier,  we  shall  have,  since 
^=18  and^2  =  8,  and  since  we  must  have 


w 


CHAIN   WITH   UNIFORM  HORIZONTAL   LOAD.  783 


but 

Xl  +  #2  =  ioo      .*.    f#2  ==  100 

Hence,  since  18  = 


2^T         3600          200 

therefore  equation  of  the  curve  is 


EXAMPLE  II.  —  Given  the  load  on  the  above  chain  as  4000 
Ibs.  per  foot  of  horizontal  length,  to  find  the  tension  at  the  low- 
est point,  also  that  at  each  end. 

Solution. 

w         i 


2H          200 


w  =  4000 


/.     2H  —  800000  .*.    H  =  400000  Ibs. 

Moreover,  load   between  lowest  point  and  highest  pier  ±= 
60  X  4000  =  240000  Ibs. 

Therefore  tension  at  highest  pier  = 


^(240000)2  -h  (400000)2  =  iooooy/(24)2  +  (40) 2 

=  10000^2176  =  466480  Ibs. 

Tension  at  lowest  pier  = 


y/(i6oooo)2  -f-  (400000)2  =  10000^256  +  1600 

=  10000^1856  =  430813  Ibs. 


784  APPLIED  MECHANICS. 

EXAMPLE  III.  —  Given  the  span  of  the  chain  as  20  feet,  and 
its  length  as  25  feet,  the  two  points  of  support  being  on  the 
same  level,  to  find  the  position  of  the  lowest  point. 

§  250.  Catenary.  —  The  catenary  is  the  form  of  the  curve 
of  a  chain,  which,  being  of  uniform  section,  is  loaded  with  its 
own  weight  only,  i.e.,  with  a  load  uniformly  distributed  along 
the  length  of  the  chain. 

To  deduce  the  equation  of  the  catenary :  if  we  assume  the 
origin,  as  before,  at  the  lowest  point  of  the  curve,  we  shall 
have  still  the  general  equation 

d2y  _  w 

but  w  in  this  case  is  not  constant. 

If  we  let  wI  =  the  load  per  unit  of  length  of  chain,  we 
shall  have 

ds 
w  =  w-i —  =  ' 

hence 

d2y       wt  ds 

~dx*~  ~H~dx 
Or,  if  we  let 


_. 
~H  ~~  ~m 

a  constant, 

#         ids 


which  is  the  differential  equation  of  the  catenary  ;  and  we  only 
need  to  integrate  it  to  obtain  the  equation  itself. 
To  do  this,  we  have 

d2y 


m\  "   '  \dxl  "    .  I        ldv\*    '  ™* 


THE   CATENARY. 


785 


therefore,  integrating,  and  observing,  that,  when  x  =  o,  ~  —  o, 

ax 

we  shall  have 


+  c. 

But,  when  x  =  o, y  —  o       .'.     c  =  —m;  hence  the  equation  is 
y=  ™J^  +  e~£)  -  m,  (3) 


and  this  is  the  equation  of  the  catenary 
when  the  origin  is  taken  at  O,  the  low- 
est point  of  the  chain. 

If  it  be  transferred  to  OIt  where  OOl 
=  m,  the  equation  becornes  (by  putting 
for  yt  y  —  m) 


O, 
FIG.  261. 


-X, 


y 


ml  * 
—  ~-[e»* 


(4) 


This  is  the  most  common  form  of  the  equation  to  the  cate- 
nary, the  origin  being  taken,  at  a  distance  below  the  lowest 

TT 

point  of   the  curve  equal  to  m  =  — ,  the  horizontal  tension 
divided  by  the  weight  per  unit  of  length  of  chain. 


;86  APPLIED   MECHANICS. 

To  find  x  in  terms  of  y,  we  have 

*-  I  2V 


**  2V    *.  M         2V     *. 

.'.      e™   -f   I   =   —e™  .%      £**  —  —  £>«  =    —  I. 

#2  /« 

Solving,  we  have 

...  f  =  loge(z±v/ZTIi 

w  (w       V  /«2         ) 

(5) 


To  find  the  length  of  the  rope  :  from  the  equation 

m(  z.         _^\ 
y  =  —  (  e™  +  e  m  J 


we  obtain 

dy       i/^        _^ 
^i  =  -(m  —      w 

dx       2 


_^\ 

<  m) 


1  1  /  —  _H£\          I  /  i  .£\ 

=y  -f^w  +  2  -|-  e  ™\  =  -f^w  +  *T»M 

/^\ 
(6) 


i  /*•*/  x.  *\  m/  x         _^\          ,  v 

.%    *     =  -  I     (^w  +  t~»*\dx  =—(e>»  —  e  ™).        (7) 

To  find  the  area  OO.A.B,  we  have 

/»*  W     ^  /   ^r  ^\  W2/  ^  _^\ 

Area  =    I    ydx  —  —  I    f  *«  +  e~>»)dx  =  —  (^»»  —  ^  «  ).  (8) 


TRANSFORMED    CATENARY.  787 


mf  *. 
arc  OB  =  —  e*"  —  e~™    =  s; 


But 


hence  area  OO.A.B  =  ms. 

This  shows,  that,  if  'the  load  should  be  distributed  in  such  a 
way  as  to  be  like  a  uniformly  thick  sheet  of  metal,  having  for 
one  side  the  catenary  and  for  the  other  the  straight  line  OVA^ 
the  equilibrium  curve  would  be  a  catenary. 

It  may  be  convenient  to  have  the  development  of  e™  and 

X 

e~™  ;  hence  they  will  be  written  here  :  — 


etc., 
m 


etc. 

m 


EXAMPLE  I.  —  Given  a  rope  90  feet  long,  spanning  a  hori- 
zontal distance  of  75  feet;  fin'd  the  equation  of  the  catenary, 
the  sag  of  the  rope,  and  the  inclination  of  the  rope  at  each 
support,  supposing  these  to  be  on  the  same  level. 

§251.  Transformed  Catenary.  —  We  have  just  seen  that 
the  catenary  is  the  form  of  chain  suited  to  a  load  which  may 
be  represented  by  a  uniformly  thick  sheet  of  metal,  with  a  hori- 
zontal extrados,  provided  the  distance  OOl  is  equal  to  m,  a  defi- 
nite quantity.  A  more  general  case,  however,  would  be  that  of 
a  chain  loaded  with  a  load  which  might  be  represented  by  a 
uniformly  thick  sheet  of  metal,  where  the  length  OOI  is  any 
given  quantity  whatever.  A  chain  so  loaded  is  called  a  trans- 
formed catenary,  and  the  catenary  itself  becomes  a  particular 
case  of  the  transformed  catenary. 

We  may  deduce  its  equation  as  follows  :  — 


788 


APPLIED  MECHANICS. 


Let  the  chain   be  represented   by  ACB,  and  let   it  be  so 
loaded  that  the  load  on  CD  is  repre- 
sented by  w  times  area  OCDE,  so  that 
B     w  =  weight  per  unit  of  area ;  then  we 
shall  have,  for  this  load, 


/*=  w 


1  I    ydx. 

t/o 


Hence,  from  what  we  have  already  seen, 

dy  _P  _ 
dx  =  ~H  = 


d2y       w 
•'•    d^  =  Hy 

Hence,  integrating,  we  have 


dy  d2y 

__      _ 

dx  dx2 


w   dy 

~~TT  y~y 

HJdx 


dy 
But,  when  -=-  =  o,  y  =  a ; 


w 


H 


Or,  if  we  write,  for  brevity,  —  =  ^2,  we  have 


dx 


2  _  y2  —  a2 
m2 


dy 


dx~  m 


dy 


dx 


—  a3 


•••  iogo>  +  y  - 


LINEAR  ARCH.  789 


But,  when  x  =  o,  y  =  a  ; 


(y  +  \y2  -  a2)  \      x 


2V  x  x  ax  -£.x 

•      —  =  &*  4-  *'**  •      v  sss  -O^T  -4-  *   w)> 

a  *       2V 

which  is  the  equation  of  the  transformed  catenary.  This 
becomes  the  catenary  itself  whenever  a  =  m. 

EXAMPLE.  —  Given  a  chain  loaded  so  that  the  load  on  CD  is 
proportional  to  the  area  OEDC.  Let  OC  =  5  feet,  £>F  =  8  feet, 
OF  —  4  feet ;  woight  per  unit  of  area  =  80  Ibs.  Find  the 
equation  of  the  transformed  catenary,  also  the  tension  at  C  and 
that  at  B. 

§  252.  Linear  Arch.  —  In  all  the  preceding  cases,  the  chain 
or  cord  is  called  upon  to  resist  a  tensile  stress  arising  from  a 
load  that  is  hung  upon  it.  If,  now,  the  cord  be  inverted,  we 
have  the  proper  equilibrium  curve  for  a  load  placed  upon  it,  dis- 
tributed in  the  same  manner  as  before ;  only  in  this  latter  case 
the  cord  would  be  subjected  to  direct  compression  throughout 
its  whole  extent.  The  equilibrium  curve  is,  then,  sometimes 
called  a  linear  arch.  The  general  equation  of  the  equilibrium 
curve  remains  just  as  before, 

d*v       w 


the  axes  being  so  chosen  that  OX  is  horizontal  and  O  Y  verti- 
cal. 

Thus,  if  it  were  required  to  find  the  form  of  the  equilibrium 
curve  or  linear  arch,  with  the  upper  boundary  of  the  loading 
horizontal,  we  should  obtain  a  transformed  catenary. 


790  APPLIED   MECHANICS. 

§  253.  Arches.  —  In  the  case  of  arches  composed  of  a  series 
of  blocks,  as  in  stone  or  brick  arches,  the  mathematical  treat- 
ment generally  used  for  determining  the  proper  form  and 
proportions  of  the  arch  has  been  quite  different  from  that  used 
for  the  determination  of  the  proper  form  and  proportions  of 
the  iron  arch,  whether  made  in  one  piece,  or  two  pieces  hinged 
together,  or  of  a  lattice. 

In  the  case  of  the  iron  arch,  the  treatment  involves  neces- 
sarily a  determination  of  the  stresses  acting  in  all  its  parts,  and 
an  adaptation  of  its  form  and  dimensions  to  the  load,  so  that  at 
no  point  shall  the  stress  exceed  the  working-strength  of  the 
material. 

In  the  case  of  the  stone  arch,  it  is  still  a  question  under 
discussion  whether  it  would  not  be  best  to  adopt  the  same 
method,  although  it  would  lead  to  a  great  deal  of  complexity,  on 
account  of  the  joints. 

Nevertheless,  the  question  usually  raised  is  one  merely  of 
stability  ;  i.e.,  as  to  the  proper  form  and  dimensions  to  pre- 
vent, not  the  crushing  of  the  stone,  though  this  must  also  be 
taken  into  account  if  there  is  any  danger  of  exceeding  it,  but 
more  especially  the  overturning  about  some  of  the  joints. 

The  question  of  the  stability  of  the  stone  arch  may  present 
itself  in  either  of  the  two  following  ways  :  — 

i°.  Given  the  arch  and  its  load,  to  determine  whether  it  is 
stable  or  not. 

2°.  Given  the  distribution  of  the  load,  to  determine  the 
suitable  equilibrium  curve,  and  hence  the  form  of  arch,  suited 
to  bear  the  given  load  with  the  greatest  economy  of  material. 

§  254.  Modes  of  giving  Way  of  Stone  Arches.  —  An  arch 
may  yield,  (i°)  by  the  crushing  of  the  stone,  (2°)  by  sliding  of 
the  joints,  (3°)  by  overturning  around  a  joint.  The  following 
figures  show  the  modes  of  giving  way  of  an  arch  by  the  last 
two  methods.  The  first  two  show  the  dislocation  of  the  arch 
by  the  slipping  of  the  voussoirs.  In  the  former  case  the 


FRICTION. 


791 


haunches  of  the  arch  slide  out,  and  the  crown  slips  down  ;  in 
the  other  case  the  reverse  happens.  The  second  two  figures 
show  the  two  methods  by  which  an  arch  may  give  way  by 
rotation  of  the  voussoirs  around  the  joints. 


FIG.  263. 


FIG.  264. 


FIG.  265. 


FIG.  266. 


Before  proceeding  farther  with  the  problem  of  the  arch,  two 
or  three  matters  of  a  more  general  nature  will  be  treated, 
which  will  be  necessary  in  its  discussion. 

§  255.  Friction.  —  Let  AB  be  a  plane  inclined  to  the  hori- 
zon at  an  angle  6.  Let  D  be  a  body  resting 
on  the  plane,  of  weight  DG  =  W.  Resolve 
W  into  two  components,  DE  and  DF  respec-  cl 
tively,  perpendicular  and  parallel  to  the 
plane.  The  component  DE  =  IV  cos  6  is 
entirely  neutralized  by  the  re-action  of  the  plane ;  while  DF 
—  Ws'm  6,  on  the  other  hand,  is  the  only  force  tending  to  make 
the  body  slide  down  the  plane.  It  is  an  experimental  fact,  that 
when  the  angle  0  is  less  than  a  certain  angle  </>,  called  the 
angle  of  repose,  the  body  does  not  slide ;  when  6  =  <£,  the  body 
is  just  on  the  point  of  sliding ;  and  when  0  is  greater  than  <£, 
the  body  slides  down  the  plane  with  an  accelerated  motion, 
showing  that  in  this  case  an  unbalanced  force  is  acting.  This 


APPLIED   MECHANICS. 


angle  <f>  depends  upon  the  nature  of  the  material  of  the  plane 
and  of  the  body,  and  on  the  nature  of  the  surfaces.  Hence, 
in  the  first  and  second  cases,  the  friction  actually  developed 
by  the  normal  pressure  DE  just  balances  the  tangential  com- 
ponent DF  ;  whereas,  in  the  third  case,  when  the  angle  of 
inclination  of  the  plane  to  the  horizon  is  greater  than  </>,  the 
tangential  component  DF  is  only  partially  balanced  by  the 
friction. 

Let  ab  be  the  plane  when  inclined  to  the  horizon  at  an 
angle  <£.  The  body  is  then  just  on  the 
point  of  sliding,  hence  the  component 
df  —  £Fsin</>  is  just  equal  to  the  fric- 
tion developed  between  the  two  surfaces. 
Moreover,  if  we  represent  by  N  the 
normal  pressure  de  =  IV  cos  <£  on  the  plane,  we  shall  have 

df  =  .Ytan<£. 

Now,  it  is  an  experimental  fact,  that  the  friction  developed 
between  two  given  surfaces  depends  only  on  the  normal  press- 
ure, i.e.,  that  the  friction  bears  a  constant  ratio  to  the  normal 
pressure  ;  and  since,  in  this  case,  the  friction  just  balances  the 
tangential  component  df  =  A^tan  <£,  the  friction  due  to  the 
normal  pressure  ./Vis 


Now,  it  makes  no  difference  what  be  the  position  of  the 
plane  surface  :  if  a  normal  pressure  N  be  exerted,  the  friction 
that  is  capable  of  being  exerted  to  resist  any  force  F  tangential 
to  the  plane,  tending  to  make  the  bodies  slide  upon  each  other, 
is  N  tan  <£  ;  and  if  the  force  F  is  greater  than  TVtan  <£,  the  bodies 
will  slide,  but  if  Fis  less  than  .W  tan  <£,  they  will  not  slide.  The 
quantity  tan  </>  is  called  the  co-efficient  of  friction,  and  will  be 
denoted  by  /. 


STABILITY  OF  BUTTRESS  ABOUT  A    PLANE  JOINT.     793 

From  the  preceding  it  is  evident,  that,  if  the  resultant  press- 
ure on  the  body  makes  with  the  normal  to  the  plane  an  angle 
less  than  the  angle  of  repose,  the  sliding  will  not  take  place ; 
whereas,  if  the  resultant  force  makes  with  the  normal  to  that 
plane  an  angle  greater  than  the  angle  of  repose,  the  body  will 
slide. 

§  256.  Stability  of  Position. — To  determine  under  what 
conditions    the    stability   of    the    block 
DGHFis  secure  against  turning  around    A 
the   edge   D:    if    the   resultant   of    the 
weight  of   the   block   and   the   pressure 
thereon  pass  outside  the  edge  D,  as  ORIt    ( 
then  the  block  will  overturn  ;   the  mo-  \Ri 

ment  of  the  couple  tending  to  overturn  it 
being  OR,  X  DE.  If,  on  the  other 
hand,  it  pass  within  the  edge,  as  OR2,  the  block  will  not  over- 
turn, since  the  force  has  a  tendency  to  turn  it  the  opposite 
way  around  D.  Hence,  in  order  that  a  block  may  not  overturn 
around  an  edge  at  a  plane  joint,  the  resultant  pressure  must 
cut  the  joint  within  the  joint  itself. 

In  any  structure  composed  of  blocks  united  at  plane  joints, 
we  must  have  both  stability  of  position  ond  stability  of  friction 
at  each  joint,  in  order  that  the  structure  may  not  give  way. 

§257.  Stability  of  a  Buttress  about  a  Plane  Joint. — 
Let  DCEF  be  a  vertical  section  of  a  buttress,  against  which 
a  strut  rib  or  piece  of  framework  abuts,  exerting  a  thrust 
P  =  ZX  =  OR.  In  order  that  the  buttress  may  not  give 
way,  it  must  fulfil  the  conditions  of  stability  at  each  joint.  Let 
AB  be  a  joint.  Should  several  pressures  act  against  the  but- 
tress, the  force  P  in  the  line  ZO  may  be  taken  to  represent 
the  resultant  of  all  the  thrusts  which  act  on  the  buttress  above 
the  joint  AB.  Let  G  be  the  centre  of  gravity  of  the  part 
ABEF,  and  let  W  —  OL  be  the  weight  of  that  part  of  the 
buttress.  Let  0  be  the  point  of  intersection  of  the  line  of 


794 


APPLIED   MECHANICS. 


direction  of  the  thrust,  and  of  the  weight  W.  Draw  the  paral- 
lelogram ORNL.  Then  will  ON  be  the  resultant  pressure  on 
the  joint  AB  :  and  the  conditions  of  stability  require  that  the 
resultant  pressure  should  cut  the  joint 
AB  at  some  point  between  A  and  B,  and 
that  its  line  of  direction  should  make 
with  the  normal  to  AB  an  angle  less 


than   the   angle   of   repose, 


and,   in 


order  that  the  buttress  may  not  give  way, 
these  conditions  must  be  fulfilled  at  each 
and  every  joint. 

Another  way  of  expressing  this  con- 
dition is  as  follows  :  The  force  tending 
to  overturn  the  upper  part  of  the  but- 
tress around  A  is  the  force  F  —  OR  ; 
and  its  moment  around  A  is  F(Ap]  =  Fp 
if  we  let  Ap  =  p,  whereas  the  moment  of  the  weight  which 
resists  this  is  W(AS)  =  Wq  if  we  let  AS  =  q.  Now,  when 
ON  passes  through  A,  we  have  Fp  =  Wq  ;  when  ON  passes 
inside  of  A,  we  have  Wq  >  Fp  ;  when  ON  passes  outside  of  A, 
we  have  Wq  <  Fp.  Hence  the  conditions  of  stability  require 
that 

Fp^.  Wq. 


or 


EXAMPLE.  —  Given  a  rectangular  buttress 
8  feet  high,  i  foot  wide,  and  4  feet  thick ;  the 
weight  of  the  material  being  100  Ibs.  per 
cubic  foot,  the  buttress  being  composed  of  8 
rectangular  blocks  1X4X1  foot.  On  this 
buttress  is  a  load  of  500  Ibs.,  whose  weight 
acts  through  K,  where  OK  =  3  feet.  Find 
the  greatest  horizontal  pressure  P  that  can  be  applied  along 
the  line  OK,  consistent  with  stability,  against  overturning 
around  each  of  the  edges  a,  b,  c,  d,  e,f,  g,  h. 


FIG.  271. 


LINE   OF  RESISTANCE   IN  A   STONE   ARCH.  795 

Solution.  —  The   weight   of   each    block   will   be   400   Ibs. 
we  shall  have  the  following  equations  :  — 


1500   -f  400    X    2 

Stability  about  a,  max  P  =  -  -    =  2300. 


=  I550. 


15°° 


O 

1600    X    2 

"     d,       "     --*  —      -  =1175- 


'50° 


„      =  1500  +  .4°°  X  2 


x 


987. 


The  least  of  these  being  987  Ibs.,  it  follows  that  the  great- 
est pressure  consistent  with  stability  against  overturning  is 
987  Ibs. 

§  258.  Line  of  Resistance  in  a  Stone  Arch.  —  In  order 
to  solve  any  problem  involving  the  stability  of  a  stone  arch,  it 
is  necessary  that  the  student  should  be  able  to  draw  a  line  of 
resistance.  To  make  plain  the  meaning  of  the  term,  the  follow- 
ing solution  of  an  example  is  given.  The  method  of  drawing 
the  line  of  resistance  employed  in  this  solution  is  given  purely 
for  purposes  of  illustration,  and  is  not  recommended  for  use  in 
practice,  as  a  suitable  method  will  be  given  later. 


796 


APPLIED   MECHANICS. 


FIG.  272. 


EXAMPLE.  —  Given  three  blocks  of  stone  of  the  form  shown 

in  the  figure  (Fig.  272),  their 
common  thickness  (perpendic- 
ular to  the  plane  of  the  paper) 
being  such  that  the  weight 
per  square  inch  of  area  (in  the 
plane  of  the  paper)  is  just  one 
pound. 

Given   AC  =   13    inches, 
BC  =    8    inches.      Suppose 
these  three  blocks  te  be  kept 
from  overturning    by  a   hori- 
zontal  force   applied    at    the 
middle  of  DE.    Find  the  least 
value  of   this  horizontal  force  consistent  with    stability  about 
the  inner  joints,  also  its  greatest  value  consistent  with  stabil- 
ity about  the  outer  joints. 
Solution. 

BK  =  i6sin  15°  =  4.14112. 
AH  =  26  sin  15°  =  6.72932. 


4.8296. 
26.25  sq.  in. 
26.25  Ibs. 

1.325. 
4.253. 
0.380. 
5.325. 
0.692. 
1.983. 
8.583. 
3.950. 


Altitude  of  each  trapezoid  =  5  cos  15° 
Area  of  each  trapezoid  =  J-jp  sin  3°° 
Weight  of  each  stone 

GG2  =  8  sin  30°  —  10.7  cos  15°  sin  15 
KK2  =8  cos  30°—  10.7  cos  15°  sin  15° 
KK^  =  10.7  cos  1  5°  cos  45°  —  8  cos  30° 
BN2   =  8  —  10.  7  cos  15°  sin  15° 
^  —  8  —  10.7  cos  15°  cos  45° 
=  10.7  cos2  15°  -  8 
=  1300330°—  10.7  cos  15°  sin  15° 
=  1300330°—  10.7  cos  15°  cos  45° 


° 


LINE   OF  RESISTANCE  IN  A   STONE   ARCH.  797 

AN*   =  13  —  10.7  cos  15°  sin  15°  =  10.325. 

AN^   =13—  10.7  cos  15°  cos  45°  =  5.692. 

AN4   =  13  —  10.7  cos2  15°  =  3.017. 

6!IJ/=  10.5  —  8  cos  30°  =  3.5  72. 

KI  M  =  10.5  —  8  sin  30°  =  6.500. 

CM   —  10.5  =  10.500. 

HI  M  =  10.5  —  13  sin  30°  =  4.000. 

Let  us  represent  the  thrust  at  Mby  T.  Then,  to  find  what 
is  the  thrust  required  to  produce  equilibrium  about  G,  we  take 
moments  about  G,  and  likewise  for  the  other  joints.  We  may 
proceed  as  follows  :  — 

INNER  JOINTS. 

Stability  about  G, 

T(G,M)  =  (26.25)  (GG2) 
or 

^(3-572)  =  (26.25)  (1.325)  •••     T=     9-74. 

Stability  about  K, 

T(K,M)  =  (26.25)  (KK2  -  KK^ 
or 

7X6.500)  =  (26.25)  (4.253  -  0.380)  .-.     T=   15.64. 

Stability  about  B, 

T(CM}    =  (26.25)  (BN*  +  B^i  -  ^0      .'•     T=    10.08 


OUTER  JOINTS. 

Stability  about  //", 

T(HiM)  =  (26.25)  (HH2  +  HHZ)  /.     T=  82.25. 

Stability  about  A, 

T(CM)    =  (26.25)  (AN2  +  AN,  +  AN,)      >.     r=  47-59 

It  is  plain,  therefore,  that,  in  order  to  have  equilibrium,  the 
A 


798 


APPLIED   MECHANICS. 


thrust  at  M  must  be  between   15.64  Ibs.  and  47.59  Ibs.  :  for, 
if  it  is  less  than   15.64  Ibs.,  the  arch  will  turn  about  an  inner 

joint;  and  if  it  is  greater  than 
47.59  Ibs.,  it  will  turn  around 
an  outer  joint. 

If,  now,  we  draw  through 
M  a  horizontal  line  to  meet 
the  vertical  drawn  through  the 
centre  of  gravity  of  the  first 
stone,  and  lay  off  a/3  =  15.64, 
and  ay  =  26.25,  then  will  the 
resultant  of  this  thrust  a/3  and 
the  weight  of  the  first  stone  ay 
be  aS;  this  being  the  resultant 
pressure  on  the  joint  FG,  its 
point  of  application  being  e. 
Next,  prolong  this  line  a8  to 
meet  the  vertical  through  the 
centre  of  gravity  of  the  second 
stone,  and  combine  aS  with  the 
weight  of  the  second  stone, 
thus  obtaining,  as  resultant 
pressure  on  the  joint  KH,  the 
force  £9,  whose  point  of  appli- 
cation is  at  K.  Compounding, 


FIG.  373. 


now,  £17  with  the  weight  of  the 
third  stone,  we  obtain,  as  final 
resultant  pressure  on  AB,  the 
force  A/x  applied  at  p.  Now, 
joining  MeKp  by  a  broken  line,  we  have  the  Line  of  Resistance 
corresponding  to  the  thrust  15.64,  or  the  minimum  horizontal 
thrust  at  M.  If,  now,  we  construct  a  line  of  resistance  with 
47.59  Ibs.,  we  obtain  the  line  Mo-<f>A,  corresponding  to  maximum 
horizontal  thrust  at  M. 


SYMMETRICAL   DISTRIBUTION  OF   THE  LOAD.  799 

If  the  arch  is  in  equilibrium,  and  if  the  horizontal  thrust  is 
applied  at  M,  it  is  plain  that  the  actual  thrust  would  either  be 
one, of  these  two  or  else  somewhere  between  these  two,  and 
hence,  that,  if  the  requisite  thrust  is  furnished  at  M  to  keep 
the  arch  in  equilibrium,  the  true  line  of  resistance  cannot  lie 
outside  of  these  two  ;  viz.,  the  line  corresponding  to  maximum 
and  that  corresponding  to  minimum  horizontal  thrust  at  M. 

If  the  separate  stones  supported  loads,  it  would  be  neces- 
sary to  take  into  account  these  loads,  in  addition  to  the  weights 
of  the  stones,  in  determining  the  horizontal  thrust,  and  drawing 
the  lines  of  resistance. 

§  259.  Arches  -with  Symmetrical  Distribution  of  the 
Load.  —  Before  considering  the  conditions  of 
stability  of  an  arch,  we  shall  proceed  to  some 
propositions  about  lines  of  resistance  corre- 
sponding to  maximum  and  minimum  horizon- 
tal thrust.  If,  in  an  arch,  we  draw  a  line  of 
resistance  AB  through  the  point  A  of  the 
crown,  and  then,  by  changing  the  horizontal 
thrust,  we  change  the  line  of  resistance  con- 
tinuously till  it  touches  the  extrados  of  the  arch  at  C',  we 
shall  evidently  have,  in  the  line  AC'S',  a  line  of  resistance 
which  has  the  greatest  horizontal  thrust  of  any  line  that  passes 
through  A,  and  lies  wholly  within  the  arch-ring.  If,  on  the 
other  hand,  we  decrease  gradually  the  horizontal  thrust  until 
the  line  touches  the  intrados  at  D ',  then  we  have  in  this  line 
the  line  of  minimum  horizontal  thrust  that  passes  through  A. 
By  lowering  the  point  A,  however,  and  keeping  the  point  C  the 
same,  we  should  obtain  new  lines  of  resistance  with  greater 
and  greater  horizontal  thrust ;  the  greatest  being  attained  when 
the  line  comes  to  have  one  point  in  common  with  the  intrados. 
Hence  a  line  of  maximum  horizontal  thrust  will  have  one  point 
in  common  with  the  extrados  and  one  point  in  common  with 
the  intrados,  the  latter  being  above  the  former. 


800  APPLIED   MECHANICS. 

On  the  other  hand,  by  retaining  the  point  D'  the  same,  and 
raising  the  point  A,  we  should  decrease  the  horizontal  thrust, 
and  thus  obtain  lines  of  resistance  with  less  and  less  horizqntal 
thrust ;  the  least  being  attained  when  the  line  of  resistance 
comes  to  have  a  point  in  common  with  the  extrados.  Hence 
the  minimum  line  of  resistance  has  a  point  in  common  with  the 
extrados  and  one  in  common  with  the  intrados,  the  latter  being 
below  the  former. 

These  cases  are  exhibited  in  the  following  figures  :  — 

i  minimum 
^maximum 


Fie.  275.  FIG.  276. 

§  260.  Conditions  of  Stability.  —  The  question  of  the  sta- 
bility of  an  arch  must  depend  upon  the  position  of  its  true 
line  of  resistance.  If  this  true  line  of  resistance  lies  within 
the  arch-ring,  the  arch  will  be  stable  provided  the  material 
of  which  it  is  made  is  incompressible.  If  this  is  not  the  case, 
the  stability  of  the  arch  will  depend  upon  how  near  the  true 
line  of  resistance  approaches  the  edge  of  the  joints ;  for  the 
nearer  it  approaches  the  edge  of  a  joint,  the  greater  the  inten- 
sity of  the  compressive  stress  at  that  joint,  and  the  greater  the 
danger  that  the  crushing-strength  of  the  stone  will  be  exceeded 
at  that  joint.  Thus,  if  the  true  line  of  resistance  cuts  any 
given  joint  at  its  centre  of  gravity,  the  stress  upon  that  joint 
will  be  uniformly  distributed  over  the  joint.  If,  however,  it 
cuts  the  joint  to  one  side  of  its  centre  of  gravity,  the  intensity 
of  the  stress  will  be  greater  on  that  side  than  on  the  opposite 
side ;  and,  if  it  is  carried  far  enough  to  one  side,  we  may  even 
have  tension  on  the  other  side. 


CRITERION  OF  SAFETY  FOR  AN  ARCH. 


801 


§  261.  Criterion  of  Safety  for  an  Arch.  —  There  are  two 
criteria  of  safety  for  an  arch,  that  have  been  used : — 

i°.  That  the  line  of  resistance  should  cut  each  joint  within 
such  limits  that  the  crushing-strength  of  the  stone  should  not 
be  exceeded  by  the  stress  on  any  part  of  the  joint. 

2°.  That,  inasmuch  as  the  joint  is  not  suited  to  bear  tension 
at  any  point,  there  should  be  no  tension  to  resist. 

The  distribution  of  the  stress  is  assumed  to  be  uniformly 
varying  from  some  line  in  the  plane  of  the  joint.  The  three 
following  figures  will,  on  this  supposition,  represent  the  three 
cases  :  — 

i°.  When  the  stress  is  wholly  compression. 

2°.  When  the  stress  becomes  zero  at  the  edge  B. 

3°.  When  the  stress  becomes  negative  or  tensile  at  B. 


R, 


R  O 
FIG.  278. 


R         O 

FIG.  279. 


In  all  three  figures,  AB  represents  the  joint  which  is  as- 
sumed to  be  rectangular  in  section,  AD  represents  the  intensity 
of  the  stress  at  A,  and  BE  that  of  the  stress  at  B ;  while  R  repre- 
sents the  point  of  application  of  the  resultant  stress,  RRT  rep- 
resenting that  resultant. 

PROPOSITION.  —  If  the  stress  on  a  rectangular  joint  vary 
uniformly  from  a  line  parallel  to  one  edge,  the  condition  that 
there  shall  be  no  tension  on  any  part  requires  that  the  result- 
ant of  the  compressive  stress  shall  be  limited  to  the  middle 
third  of  the  joint.  . 

PROOF.  —  Let  AB  (Fig.  278)  represent  the  projection  of 
the  joint  on  the  plane  of  the  paper.  It  is  assumed  that  the 


8O2  APPLIED  MECHANICS. 

stress  is  uniformly  varying  ;  and,  if  there  is  to  be  no  tension 
anywhere,  the  intensity  at  one  edge  must  not  have  a  value  less 
than  zero,  hence  at  the  limiting  case  the  value  must  be  zero  ; 
hence  this  limiting  case  is  correctly  represented  by  the  figure, 
and  the  resultant  of  the  compression  will  be  for  this  case  at  the 
centre  of  stress.  Thus,  if  AD  represent  the  greatest  intensity 
of  the  stress,  then  we  shall  have,  if  B  be  the  origin  and  BA  the 
axis  of  xy  if  the  axis  of  y  be  perpendicular  to  AB  at  B,  and  if 
we  let  a  =  intensity  of  stress  at  a  unit's  distance  from  B,  that 
RR,  =  affxdxdy,  and  (BR)  (RR,)  =  affx2dxdy; 


ff&dxdy 
BR  = 


if  b  =  breadth,  and  h  =  BA  =.  height  of  rectangle. 

Hence,  if  the  resultant  of  the  compression  be  nearer  A  than 
R,  there  will  be  tension  at  B ;  and,  on  the  other  hand,  if  it  be 
nearer  B  than  \h,  there  will  be  tension  at  A.  Hence  follows 
the  proposition  as  already  stated. 

While  the  above  is  probably  the  condition  most  generally 
used  to  determine  the  stability  of  an  arch,  at  the  same  time,  if 
there  is  any  danger  that  the  intensity  of  the  stress  at  any  part 
of  any  joint  may  exceed  the' working  compressive  strength  of 
the  stone,  this  ought  to  be  examined,  and  hence  a  formula  by 
which  it  may  be  done  will  be  deduced. 

Let  AB  (Fig.  279)  be  the  joint,  and  let,  as  before,  b  be  its 
breadth,  and  h  =  AB  =  depth  ;  then,  suppose  the  pressure  to 
be  uniformly  varying,  DA  =f=  the  working-strength  per  unit 
of  area  =  greatest  allowable  intensity  of  compression  ;  then  the 
entire  stress  on  the  joint  will  be  represented  by  the  triangle 
A  CD,  for  the  joint  is  incapable  of  resisting  tension. 
Hence 

=  \AC  .'. 


POSITION  OF  THE    TRUE  LINE   OF  RESISTANCE.       803 


but 


and  this  is  the  least  distance  from  the  outer  edge  at  which  the 
resultant  should  cut  the  joint. 

We  thus  obtain,  in  terms  of  the  pressure  on  any  joint,  and 
of  the  working-strength  of  the  material,  the  limits  within  which 
the  line  of  resistance  should  pass,  in  order  that  the  working- 
strength  of  the  stone  may  not  be  exceeded. 

§  262.  Position  of  the  True  Line  of  Resistance The 

question  of  the  most  probable  position  of  the  true  line  of 
resistance  involves  the  discussion  of  the  properties  of  the 
elastic  arch.  This  discussion  will  be  given  later ;  but,  for  the 
present,  the  statement  only  of  the  following  proposition,  due  to 
Dr.  Winkler,  will  be  given  :  — 

"  For  an  arch  of  constant  section,  that  line  of  resistance  is 
approximately  the  true  one  ivkich  lies  nearest  to  the  axis  of  the 
arch-ring,  as  determined  by  the  method  of  least  squares." 

From  this  it  will  follow  :  — 

i°.  That,  if  a  line  of  resistance  can  be  drawn  in  the  arch- 
ring,  then  the  true  line  of  resistance  will  lie  in  the  arch-ring ; 
and 

2°.  That,  if  a  line  of  resistance  can  be  drawn  within  the 
middle  third  of  the  arch-ring,  then  the  true  line  of  resistance 
will  lie  in  the  middle  third. 

But,  before  proving  this  proposition,  the  proposition  will  be 
used,  and  the  method  explained,  for  determining  whether  a  line 
of  resistance  can  be  drawn  within  the  arch-ring :  for,  if  it  can, 
then  the  true  line  of  resistance  must  lie  within  the  arch-ring ; 
and  if  no  line  of  resistance  can  be  drawn  within  the  arch-ring, 
then  the  true  Mne  of  resistance  cannot  pass  within  the  arch- 
ring,  and  the  arch  would  necessarily  be  unstable,  even  if  the 
materials  were  incompressible. 

By  following  the  same  method,  we  could  determine  whether 


804  APPLIED   MECHANICS. 

it  was  possible  to  draw  a  line  of  resistance  within  the  middle 
third  of  the  arch-ring ;  and,  if  this  is  found  to  be  possible,  we 
should  know  that  the  true  line  of  resistance  will  pass  within 
the  middle  third  of  the  arch-ring. 

Hence  our  most  usual  criterion  of  the  stability  of  a  stone 
arch  is,  whether  a  line  of  resistance  can  be  passed  within  the 
middle  third  of  the  arch-ring. 

If  the  condition  be  used,  that  the  working-strength  of  the 
stone  for  compression  be  not  exceeded,  then,  instead  of  the 
middle  third,  we  shall  have  some  other  limits. 

In  what  follows,  an  explanation  will  be  given  of  Dr.  Scheff- 
ler's  method  (that  most  commonly  employed)  of  determining 
whether  a  line  of  resistance  can  be  drawn  within  the  arch-ring, 
inasmuch  as  the  same  method  can  be  employed  to  determine 
whether  such  a  line  can  be  drawn  within  the  middle  third  or 
within  any  other  given  limits. 

§  263.  Preliminary  Proposition  referring  to  Arches  Sym- 
metrical in  Form  and  Loading.  —  An  arch  and  its  load  being 
given,  a  line  of  resistance  can  always  be  made  to  pass  through 
any  two  given  points ;  hence,  if  any  two  points  of  a  line  of 
resistance  are  given,  the  line  is  determined. 

Proof.  —  Let  the  arch  be  that  shown  in  Fig.  281  ;  and  let  us 
consider  first  the  special  case  when  the  two  given  points  are  A, 
the  top  of  the  crown-joint,  and  6*4,  the  foot  of  the  springing- 
joint.  In  this  case,  the  only  quantity  to  be  determined  is  the 
thrust  at  A.  Let  this  thrust  be  denoted  by  T;  let  P  be  the 
total  weight  of  the  half-arch  and  its  load ;  let  a  be  the  perpen- 
dicular distance  of  the  point  G4  from  a  vertical  line  through  the 
centre  of  gravity  of  the  entire  half-arch  and  its  load ;  let  h  be 
the  vertical  depth  of  G4  below  A.  Then,  taking  moments  about 


G4,  we  must  have 


Th  =  Pa 


DR.   SCHEFFLER'S   METHOD.  805 

and  the  line  of  resistance  can  then  be  drawn  with  this  thrust, 
as  has  been  done  in  the  figure.  Next  take  the  general  case, 
when  the  given  points  are  not  in  these  special  positions.  Let 
them  be  any  two  points,  as  A2  and  Gv 

In  this  case,  the  point  of  application  of  the  thrust  at  the 
crown  is  not  necessarily  A,  but  may  be  some  other  point  of  the 
crown-joint:  hence  the  quantities  to  be  determined  are  two; 
viz.,  the  thrust  T  at  the  crown,  and  the  distance  x  of  its  point 
of  application  below  A.  Let  the  combined  weight  of  the  first 
two  voussoirs  and  their  load  be  P^  and  the  horizontal  distance 
of  A2  from  a  vertical  line  through  the  centre  of  gravity  of  Pl  be 
4p 

Let  P2  be  the  combined  weight  of  the  first  three  voussoirs 
and  their  load,  and  let  a2  be  the  horizontal  distance  of  G3  from 
a  vertical  line  through  the  centre  of  gravity  of  P2. 

Let  the  vertical  depth  of  A2  below  A  be  ku  and  that  of  G3 
below  A  be  k2.  Then,  taking  moments  about  A2  and  G3  respec- 
tively, we  shall  have 

—  x)  =  P&     and     T(h2  —  x)  =  P2a2, 


two  equations  to  determine  the  two  unknown  quantities  T  and 
^r,  which  can  easily  be  solved  in  any  special  case  ;  and  the  result- 
ing line  of  resistance  can  be  drawn,  which  will  pass  through  the 
two  given  points. 

§  264.  Dr.  Scheffler's  Method  of  Determining  the  Possi- 
bility of  Passing  a  Line  of  Resistance  within  the  Arch- 
Ring.  —  In  using  Scheffler's  method  of  determining  whether  it 
is  possible  to  pass  a  line  of  resistance  within  the  arch-ring  or 
not,  we  should  proceed  as  follows  ;  viz.,  — 

First  pass  a  line  of  resistance  through  i,  the  top  of  the 
crown-joint  (Fig.  280),  and  e,  the  inside  of  the  springing-joint. 
If  this  line  lies  wholly  within  the  arch-ring,  it  proves  that  a 
line  of  resistance  can  be  drawn  within  the  arch-ring. 

If  this  line  of  resistance  does  not  pass  entirely  within  the 


806  APPLIED   MECHANICS. 

arch-ring,  proceed  as  follows  :  Suppose  the  line  thus  drawn  to 
be  \abcde,  passing  without  the  arch-ring  on  both  sides,  as 
shown  in  the  figure.  Then  from  a,  the  point 
where  it  is  farthest  from  the  extrados  of  the  arch- 
ring,  draw  a  normal  to  the  extrados,  and  find  the 
point  where  this  normal  cuts  the  extrados  :  in  this 
case,  a  is  the  point  in  question.  In  this  way 
determine  also  the  point  7,  where  the  normal 
from  d  cuts  the  intrados  ;  then  pass  a  new  line 
of  resistance  through  the  points  a  and  7,  deter- 
mining the  thrust  and  its  point  of  application.  If  this  new 
line  of  resistance  lies  within  the  arch-ring,  then  it  is  plain  that 
it  is  possible  to  draw  a  line  of  resistance  within  the  arch-ring ; 
if  not,  it  is  not  at  all  probable  that  it  is  possible  to  draw  such  a 
line. 

If  the  line  of  resistance  drawn  through  i  and  e  goes  outside 
the  arch-ring  only  beyond  the  extrados,  as  at  a,  we  should  draw 
our  second  line  of  resistance  through  a  and  e ;  if,  on  the  other 
hand,  it  goes  outside  only  below  the  intrados,  as  at  d,  we  should 
draw  our  second  line  through  I  and  7. 

In  the  construction,  we  make  use  of  a  slice  of  the  arch  in- 
cluded between  two  vertical  planes  a  unit  of  distance  apart ;  and 
we  take  for  our  unit  of  weight  the  weight  of  one  cubic  unit  of 
the  material  of  the  voussoirs,-  so  that  the  number  of  units  of 
area  in  any  portion  of  the  face  of  the  arch  shall  represent  the 
weight  of  that  portion  of  the  arch. 

We  next  draw,  above  the  arch,  a  line  (DD4  in  Fig.  281), 
straight  or  curved,  such  that  the  area  included  between  any 
portion  of  it,  as  £>s£>2,  the  two  verticals  at  the  ends  of  that  por- 
tion, and  the  extrados  of  the  arch-ring,  shall  represent  by  its 
area  the  load  upon  the  portion  of  the  arch  immediately  below 
it.  This  line  will  limit  the  load  itself  whenever  this  is  of  the 
same  material  as  the  voussoirs  ;  otherwise  it  will  not.  We  shall 
always  call  it,  however,  the  extrados  of  the  load. 


DR.  SCHEFFLER'S  METHOD. 


807 


The  mode  of  procedure  will  best  be  made  plain  by  the  solu- 
tion of  examples ;  and  two  will  be  taken,  in  the  first  of  which 
only  one  trial  is  necessary  to  construct  a  line  of  resistance  that 
shall  lie  wholly  within  the  arch-ring,  and,  in  the  second,  two 
trials  are  necessary. 

EXAMPLE.  —  The  half-arch  under  consideration  is  shown  in 
Fig.  281,  GG4  being  the  intrados,  AA4  the  extrados  of  the  arch, 


FIG.  281. 


and  DD4  the  extrados  of  the  load.     The  arcs  GG4  and  AA4  are 
concentric  circular  arcs.     The  data  are  as  follows  :  — 

Span  =  2(G4O)  =  6.00  feet, 

Rise  =  GO         =0.50  foot, 

Thickness  of  voussoirs  =  AG  —  A4G4       =0.75  foot, 

Height  of  extrados  of  load  above  A  —  AD         =  0.80  foot. 

The  position  of  the  joints  is  not  assumed  to  be  located.  We 
therefore  draw  through  A  a  horizontal  line  AB,  and  divide  this 
into  lengths  nearly  equal,  unless,  as  is  usual  near  the  springing, 
there  is  special  reason  to  the  contrary.  Thus,  we  make  the  first 
three  lengths  each  equal  to  i  foot,  and  thus  reach  a  vertical 


8o8 


APPLIED   MECHANICS. 


through  £4;  and  then  the  last  division  has  a  length  of  0.24  foot. 
We  have  thus  divided  the  half-arch  and  its  load  into  four  parts  ; 
viz.,  GDD,H»  H,D,DJIM  H2D2DzG,y  and  G4D3D4A4,  the  loads 
on  these  respective  portions  being  represented  by  their  areas 
respectively.  We  assume  the  centre  of  gravity  of  each  load  to 
lie  on  its  middle  vertical ;  and  we  then  proceed  to  determine  the 
numerical  values  of  the  several  loads,  the  distances  of  their 
centres  of  gravity  from  a  vertical  through  the  crown,  also  the 
amount  and  centre  of  gravity  of  the  first  and  second  loads 
together,  then  of  the  first,  second,  and  third,  etc. 

The  work  for  this  purpose  is  arranged  as  follows  :  — 


(1) 

(2) 

(3) 

(4) 

(6) 

(6) 

(7) 

(8) 

(9) 

(10) 

•s-s 

11 

Length. 

Height. 

Area. 

Lever 
Arm. 

Moment. 

Partial  Sums. 

Area. 

Moment. 

Lever 

Arm. 

j 

I.OO 

i-57 

1-570 

0.50 

0.785 

I 

1-570 

0.785 

0.500 

2 

I.OO 

1.68 

1.  680 

I.50 

2.520 

1  +  2 

3-250 

3-305 

I.OI7 

3 

I.OO 

1.90 

1.900 

2.50 

4-75° 

1  +  2+3 

5-I50 

8.055 

I-563 

4 

0.24 

1.72 

0.413 

3.12 

1.287 

1+2+3+4 

5-563 

9-342 

1.680 

- 

3-24 

- 

5-563 

- 

9-342 

- 

- 

- 

- 

Column  (i)  shows  the  number  of  the  voussoir. 

"  (2)  gives  the  horizontal  lengths  of  the  several  trape- 
zoids. 

"        (3)  gives  the  middle  heights  of  the  trapezoids. 

(4)  gives  the  areas  of  the  trapezoids,  and  is  obtained  by 
multiplying  together  the  numbers  in  (2)  and  (3). 

"  (5)  gives  the  distances  from  A  to  the  middle  lines  of 
the  trapezoids. 

"  (6)  gives  the  products  of  (4)  and  (5),  giving  the  moments 
of  the  respective  loads  about  an  axis  through  A 
perpendicular  to  the  plane  of  the  paper. 


DR.   SCHEFFLER'S   METHOD. 


809 


Column  (7)  merely  indicates  the  successive   combinations   of 

voussoirs. 
"        (8)  has  for  its  numbers,  — 

i°.  The  area  representing  the  first  load. 
2°.  The  area  representing  the  first  two  loads. 
3°.  The  area  representing  the  first  three. 
4°.  The  area  representing  the  first  four. 
"        (9)  has  for  its  numbers,  — 

i°.  The  moment  of  the  first  load  about  A. 
2°.  The  moment  of  the  first  and  second  loads 

about  A. 
3°.  The  moment  of  the  first,  second,  and  third 

loads  about  A. 
4°.  The  moment  of  the  first,  second,  third,  and 

fourth  loads  about  A. 

(10)  is  obtained  by  dividing  column  (9)  by  column  (8) ; 
the  quotients  being  respectively  the  distance 
'from  A  to  the  centres  of  gravity  of  the  first,  of 
the  first  and  second,  of  the  first,  second,  and 
third,  and  of  the  first,  second,  third,  and  fourth 
loads. 

The  calculation  thus  far  is  purely  mathematical,  and  merely 
furnishes  us  with  the  loads  and  their  points  of  application ;  in 
other  words,  furnishes  us  the  data  with  which  to  begin  our 
calculation  of  the  thrust.  Before  passing  to  this,  it  should  be 
said,  however,  that  we  now  assume  the  joints  to  be  drawn 
through  the  points  AIt  A2,  A3,  and  A4,  and  generally  normal  to 
the  extrados  of  the  arch. 

In  this  proceeding,  we,  of  course,  make  an  error  which  is 
very  small  near  the  crown  and  increases  near  the  springing  of 
the  arch  ;  this  error,  in  the  case  of  voussoir  A^A2G^G2,  amounts 
to  the  difference  of  the  two  triangles  A2G2H2  and  A^G^H^.  A 
manner  of  making  a  correction  by  moving  the  joint  will  be 
•explained  later ;  but  now  we  will  complete  our  example,  as  the 


8 10  APPLIED   MECHANICS. 

errors  are  not  serious  in  this  example.  We  now  pass  a  line  of 
resistance  through  A,  the  upper  point  of  the  crown-joint,  and 
G4,  the  lower  point  of  the  springing.  For  this  purpose  take 
moments  about  G4 ;  and  we  shall  have,  if  T  =  thrust  at  the 
crown, 

1.25^=  (5-563)(3  -  1-68), 

since  5.563  is  the  whole  weight,  and  3  —  1.68  is  its  leverage 
about  Gr 

Hence 

1.257'=  (5.563)  (1.32)  =  7.343 

.-.    r=5.87. 

Hence  we  proceed  to  draw  a  line  of  resistance  through  A, 
assuming,  as  the  horizontal  thrust,  5.87.  To  do  this  we  proceed 
as  follows  :  From  a,  the  point  of  intersection  of  AD  with  the 
vertical  through  the  centre  of  gravity  of  the  first  trapezoid,  we 
lay  off  ab  to  scale  equal  to  5.87,  and  then  lay  off  bC  vertically 
to  scale  equal  to  1.57,  the  first  load ;  then  will  Ca  be  the  result- 
ant pressure  on  joint  AtGlf  and  its  point  of  application  will  be 
P,  which  gives  us  one  point  in  the  line  of  resistance.  To 
obtain  the  point  Plt  we  lay  off  AaI  =  1.017,  the  lever  arm  of 
the  first  two  loads  ;  then  lay  off  aldI  —  5.87,  the  thrust ;  then 
lay  off  b£i  equal  to  3.25,  the  weight  of  the  first  two  loads.. 
Then  will  £",#,  be  the  pressure  on  the  second  joint ;  and  the 
point  PI9  or  its  point  of  application,  is  at  the  intersection  of 
Cjfli  with  A2G2. 

Then  lay  off  Aa2  =  1.563,  azb2  =  5.87,  b2C2  =  5.150;  and 
Pm  the  next  point  of  the  line  of  resistance,  is  the  intersection 
of  C2a2  with  A3G3.  Then  lay  off  Aa3  =  1.680,  ajb^  =  5.87, 
bjC^  =  5.563  ;  and  C2a3  is  the  pressure  on  the  springing,  and 
this  will  intersect  A4G4  at  G4  unless  some  mistake  has  been 
made  in  the  work.  Then  is  APP^P2G4  the  line  of  resistance 
through  A  and  G4,  and  this  lies  entirely  within  the  arch-ring. 


SCHEFFLER'S  MODE   OF  CORRECTING    THE  JOINTS.     8ll 

Hence  we  conclude  that  it  is  possible  to  draw  a  line  of  resist- 
ance within  the  arch-ring  without  having  recourse  to  another 
trial. 

§  265.  Scheffler's  Mode  of  Correcting  the  Joints.  —  The 
following  is  the  approximate  construction  given 
by  Scheffler  for  correcting  the  joint :  Let  DCG 
be  the  side  of  the  trapezoid,  and  CH  the  uncor- 
rected  joint.  From  b,  the  middle  point  of  GH, 
draw  Db ;  then  draw  Gc  parallel  to  bD,  and  ch 
parallel  to  CH.  Then  will  ch  be  the  corrected 
joint. 

Conversely,  having  given  the 
joint  CH,  to  find  the  side  of  the  trapezoid  which 
limits  the  portion  of  the  load  upon  it :  through 
C  draw  DG  vertical ;  join  D  with  b,  the  middle 
point  of  GH ;  then  draw  Cg  parallel  to  Db ; 
G  then,  from  g,  drawing  dg  vertical,  we  thus  have 

FIG.  283.          the  Desired  side  of  the  trapezoid. 
§  266.  Another   Example.  — Another  example  will  now  be 
solved,  which  necessitates  two  trials,  and  where  some  of  the 
joints  have  to  be  corrected.    It  is  practically  one  of  Scheffler's. 
The  dimensions  of  the  arch  are  as  follows  :  — 

Half-span 32-97  feet. 

Rise 24.74  feet. 

Thickness  of  ring 5.15  feet. 

Height  of  load  at  crown 8.24  feet. 

Height  of  load  at  springing 33-5<>  feet. 

The  arch  may  be  drawn  by  using,  for  the  intrados,  two- 
circular  arcs.  Beginning  at  the  springing,  draw  a  60°  arc  with 
a  radius  of  one-fourth  the  span  ;  then,  with  an  arc  tangent  to 
this  arc,  continue  to  the  crown,  the  proper  rise  having  been 
previously  laid  off.  The  work  for  drawing  a  line  of  resistance 


-812 


APPLIED   MECHANICS. 


through   the   top   of  the   crown-joint   and   the   inside   of   the 
springing  will  be  given  without  comment.     It  is  as  follows  :  — 


(1) 

(2) 

(3) 

(4) 

(5) 

(6) 

(7) 

(8) 

(9) 

(10) 

Number  of 
the  Voussoir. 

Length. 

Height. 

Area. 

Lever 
Arm. 

Mo- 
ment. 

Partial  Sums. 

Area. 

Moment. 

Lever 
Arm. 

2 
3 

4 
5 
6 

8.24 
8.24 
8.24 
4-13 
4-13 

5-r4 

14.1 

164 
I8.3 
22.6 

27.1 

34-7 

116.18 
I35-I4 

150.79 
93-34 
111.92 
178.36 

4.12 
12.36 
20.60 
26.79 
30.92 

35-55 

479 
1670 
3106 
2500 
346i 
6341 

I 

1+2 

I  +  ...+3 
i+.  ..+4 
H-...+5 
i+.  ..+6 

116.18 

25L32 
402.11 

495-45 
607.37 

78573 

479 
2149 
5255 
7755 
11216 

J7557 

4.12 

8.55 

I3-07 
15.65 
18.46 
22.34 

- 

- 

- 

78573 

- 

17557 

- 

- 

- 

1 

29.897*  =  (785. 73)  (32.97  -  22.34), 
29.897^  =  8352.31 

.-.     T  =  279.4. 

Hence  we  construct  the  line  of  resistance  passing  through  the 
top  of  the  crown-joint  and  the  inside  of  the  springing,  using 
the  thrust  279.4. 

The  construction  is  shown  in  the  figure,  and  is  entirely 
similar  to  that  previously  used,  with  the  single  exception  that 
the  upper,  instead  of  the  lower,  half  of  the  rectangle  is  used, 
in  each  case,  in  constructing  the  parallelogram  of  forces,  to 
determine  the  pressure  on  each  joint :  this  is  merely  a  matter 
of  convenience.  The  student  will  readily  identify  this  first 
line  of  resistance,  and  will  see  that  it  goes  outside  the  arch- 
ring  both  above  and  below,  being  farthest  above  the  extrados 
at  the  first  joint  from  the  crown,  and  farthest  inside  of  the 
intrados  opposite  the  first  joint  from  the  springing.  Hence  we 


SCHEFFLER'S  MODE   OF  CORRECTING    THE  JOINTS. 


proceed  to  pass  a  new  line  of  resistance  through  the  top  of  the 
first  joint  from  the  crown,  and  the  inside  of  the  first  joint  from 
the  springing. 


FIG.  284. 

For  this  purpose  we  do  not  need  to  make  out  a  new  table,, 
as  it  is  not  necessary  to  insert  any  new  joints.  We  need  only 
two  more  dimensions,  i.e.,  the  vertical  depth  of  each  of  these 
points  below  the  top  of  the  crown :  these  depths  are  respec- 
tively 2.38  and  21.85. 

Hence  we  proceed  as  follows  :  — 
Let  T  =  thrust  at  the  crown, 

x  =  distance  of  its  point  of  application  below  the  top  of 
the  crown-joint. 

i°.  Take  moments  about  the  upper  one  of  the  two  points* 
and  we  have 

7X2.38  -  x)  =  (n6.i8)(8.24  -  4.12)  =  478.66. 


8 14  APPLIED   MECHANICS. 

2°.  Take  moments  about  the  lower  one  of  the  two  points, 
and  we  have 

7X21.85  -  x)  =  (607.37)  (30.90  -  18.46)  =  7458.50. 

Solving  these  equations,  we  obtain 

T  =  358-9>    x  =  1-046- 

Hence  we  lay  off  on  the  crown-joint  a  distance  1.046  below  the 
top  of  the  crown-joint,  and  through  this  point  draw  a  horizontal 
line,  this  line  being  the  line  of  action  of  the  thrust.  Then, 
making  the  construction  for  a  new  line  of  resistance  just  as 
before,  only  using  this  new  point  of  application  of  the  thrust, 
and  using  for  thrust  358.9,  we  shall  obtain  a  new  line  of  resist- 
ance, which  passes  through  the  desired  points ;  and,  since  this 
line  lies  within  the  arch-ring,  we  therefore  conclude  that  it  is 
possible  to  draw  a  line  of  resistance  within  the  arch-ring. 

Another  method  of  drawing  a  line  of  resistance  frequently 
pursued  is  the  following:  — 

After  determining  the  loads  on  the  successive  voussoirs, 
and  also  the  thrust  for  the  particular  line  of  resistance  which 
we  wish  to  draw,  layoff  these  Toads  and  thrust  to  scale  in  their 
proper  order  and  directions,  and  construct  a  force  polygon 
(see  §  126),  then  construct  the  corresponding  frame  (equilib- 
rium polygon),  which  shall  have  its  apices  on  the  vertical  lines 
passing  through  the  centres  of  gravity  of  the  loads  as  drawn  in 
the  figure  of  the  arch.  The  intersections  of  the  lines  of  the 
frame  with  the  joints  give  us  points  of  the  line  of  resistance, 
and  the  line  of  resistance  can  be  drawn  by  joining  them. 

Thus,  applying  the  solution  to  Fig.  281,  we  lay  off 
ab  =  1.570,  be  =  i. 680, 

cd  =  1.900,  de  —  0.413. 

Also,  lay  off 

oa  =  5.87, 

and  draw  the  lines  ob,  ocy  od,  and  oe. 


DRAWING  A    LINE   OF  RESISTANCE. 


815 


Then  from  A  draw  Aa  parallel  to  oa,  then  draw  aa^  parallel 
to  ob,  a^2  parallel  to  oc,  dzaz  parallel  to  od,  and  a3a4  parallel  to 


8i6 


APPLIED  MECHANICS. 


oe,  this  last  prolonged  backwards  of  course  passes  through  G4  ; 
then  will  the  line  of  resistance  be  obtained  by  joining  the  points 


The  last  figure  on  preceding  page  shows  the  same  method 
applied  to  Fig.  284. 

In  these  examples  the  entire  arch-ring  has  been  used  instead 
of  the  middle  third  or  other  limits,  so  as  to  make  the  figures 
comparable  with  Figs.  281  and  284;  but  of  course  in  practice 
the  middle  third  should  be  used  instead  of  the  entire  arch- 
ring. 

§  267.  Examples.  —  Four  more  examples  will  now  be  given 
to  be  worked  out  by  the  student.  The  dimensions  are  approx- 
imately those  given  in  some  of  Scheffler's  examples. 

EXAMPLE  I.—  Half-span  ==  CD  =  65.16  feet,  rise  =  FD  — 
13.85  feet,  AF  =  5.32  feet,  AE  =  6.40  feet.  The  arcs  CF  and 
AG  are  concentric  circular  arcs.  Given  width  of  first  five 
horizontal  divisions  of  line  AB,  counting  from  A,  each  10.66 
feet;  width  of  sixth  division,  n.86  feet;  of  seventh,  2.2  feet. 
Determine  the  possibility  of  drawing  a  line  of  resistance  in  the 
arch-ring. 


FIG.  285. 


EXAMPLE  II.— Half-span  =  63.98  feet,  rise  =  FD  —  31.99 
feet,  AF  —  CG  =  5.32  feet,  AE  =  2.13  feet.  The  intrados  and 
extrados  of  this  arch  are  seven  centred  ovals,  both  drawn  from 


EXAMPLES. 


817 


the  same  centres.  Beginning  at  the  springing,  an  arc  with  a 
radius  of  21  feet  is  drawn,  subtending  39°;  the  curve  is  con- 
tinued by  a  curve  subtending  24°,  and  having  a  radius  of  35-55 
feet.  From  F  an  arc  subtending  10°  is  drawn  from  a  centre 
on  FD  produced,  and  with  a  radius  of  152  feet;  the  curve  is 
completed  by  an  arc  connecting  the  second  and  last. 


FIG.  286. 

Given  horizontal  width  of  each  of  first  six  divisions,  counting 
from  A,  10.66  feet ;  horizontal  width  of  seventh  division,  5.32 
feet.  Determine  the  possibility  of  drawing  a  line  of  resistance 
in  the  arch-ring. 


FIG.  287. 


EXAMPLE   III.  —  Given  span  =  74.18  feet;   rise   =  45.83 


8i8 


APPLIED   MECHANICS, 


feet ;  radius  of  intrados  —  82.42  feet ;  radius  of  extrados  = 
91.18  feet ;  height  of  load  at  crown  =  8.24  feet ;  width  of  each 
of  five  divisions  nearest  crown  =  8.24  feet;  width  of  sixth 
stone  =  4.13  feet.  Determine  the  possibility  of  drawing  a  line 
of  resistance  within  the  arch-ring. 

EXAMPLE  IV.  —  Given  span  =  37.07  feet ;  thickness  of 
ring  =  AB  —  3.08  feet ;  height  of  load  =  BC  =  82.42  feet. 
Determine  the  possibility  of  drawing  a  line  of  resistance  within 
the  arch-ring. 


FIG.  288. 


§  268.  Criterion  of  Stability.  —  It  has  already  been  stated, 
that,  if  a  line  of  resistance  can  be  drawn  within  the  arch-r.ng, 
then  the  true  line  of  resistance  will  lie  within  the  arch-ring. 


UNSYMMETRICAL   ARRANGEMENT.  819 

With  those  who,  like  Scheffler,  consider  the  material  of  the 
voussoirs  incompressible,  the  criterion  of  stability  of  an  arch  is, 
that  it  should  be  possible  to  draw  aline  of  resistance  within  the 
arch-ring. 

On  the  other  hand,  Rankine  would  decide  upon  the  stability 
of  an  arch  by  determining  whether  a  line  of  resistance  can  be 
drawn  within  the  middle  third  of  the  arch-ring. 

Other  limits  have  been  adopted  instead  of  the  middle  third. 
In  some  cases  the  only  reason  for  deciding  upon  what  these 
limits  should  be  has,  been  custom  or  precedent. 

They  might  also  be  determined  so  that  there  should  be  no 
danger  of  exceeding  the  crushing-strength  of  the  stone. 

It  is  needless  to  say  that  the  first  method  is  incorrect ;  for 
the  material  of  the  voussoirs  is  never  incompressible,  and  an 
arch  where  the  true  line  of  resistance  touches  the  intrados  or 
extrados  could  not  stand,  as  the  stone  would  be  crushed. 

Nevertheless,  no  example  will  be  solved  here,  where  we  de- 
termine the  possibility  of  drawing  a  line  of  resistance  within 
the  middle  third,  or  other  limits  than  the  entire  arch-ring,  as  the 
method  of  procedure  is  entirely  similar  to  what  we  have  done, 
the  computation  of  the  entire  table  being  the  same  in  all  cases, 
the  only  difference  occurring  in  the  computation  of  the  thrust 
and  its  point  of  application,  and  the  consequent  construction  of 
the  line  of  resistance.  The  method  to  be  pursued  is,  as  before, 
by  taking  moments  about  the  points  through  which  it  is  desired 
that  the  line  of  resistance  shall  pass. 

§  269  Unsymmetrical  Arrangement.  —  When  the  arch  is 
unsymmetrical,  either  in  form  or  loading,  the  same  criterion  as 
to  being  able  to  pass  a  line  of  resistance  within  the  middle  third 
or  other  limits  of  the  arch-ring  will  serve  to  determine  its  sta 
bility.  The  method  of  procedure  differs,  however,  from  the  fact, 
that  whereas  we  have  heretofore  found  it  necessary  to  study 
only  the  half-arch  and  its  load,  and  have  had  the  advantage  of 
knowing,  from  the  symmetry  of  arch  and  load,  that  the  thrust  at 


820  APPLIED   MECHANICS. 

the  crown  is  horizontal,  we  have  not  that  advantage  here,  and 
hence  we  must  study  the  entire  arch,  and  we  must  assume  that 
the  thrust  at  the  crown  may  be  oblique,  and  hence  have  a  verti- 
cal as  well  as  a  horizontal  component. 

In  this  case  it  will  be  necessary  to  have  three  instead  of  two 
points  given,  in  order  to  determine  a  line  of  resistance. 

If  we  assume  (Fig.  289)  a  vertical  joint  at  the  crown,  and  let 
P  =  vertical  component  of  the  thrust  at 
the  crown,  A  =  horizontal  component  of 
the  thrust  at  the  crown,  x  —  distance 
of  point  of  application  of  thrust  at  the 
crown  below  upper  point  of  crown-joint, 
FIG  2g  we  have  thus  three  unknown  quantities, 

and  we  shall  therefore  need  three  equa- 
tions to  determine  them. 

In  this  case,  therefore,  we  must  have  three  points  of  the  line 
of  resistance  given,  in  order  to  determine  it ;  and  a  reasoning 
similar  to  that  pursued  in  §  263  would  show  that  a  line  of  re- 
sistance can  always  be  passed  through  any  three  given  points. 

In  performing  the  work,  we  should  need  to  make  out  a  table 
for  the  part  of  the  arch  on  each  side  of  the  crown-joint,  show- 
ing the  loads,  and  centres  of  gravity  of  the  loads,  on  each  vous- 
scir,  and  on  combinations  of  the  first  two,  first  three,  etc. ;  this 
portion  of  the  work  being  entirely  similar  to  that  done  in  the 
case  of  arches  of  symmetrical  form  and  loading,  only  that  we 
require  a  separate  table  for  the  parts  on  each  side  of  the  crown- 
joint. 

When  these  two  tables  have  been  worked  out,  we  next  pro- 
ceed to  impose  the  conditions  of  equilibrium  by  taking  moments 
about  each  of  the  three  points  given. 

Thus,  suppose  that  (as  is  usually  done  first)  we  pass  a  line 
of  resistance  through  the  top  of  the  crown-joint  and  the  inside 
of  each  springing-joint,  we  then  have  only  two  unknown  quan- 
tities to  determine  ;  viz.,  P  and  <2,  inasmuch  as  x  becomes  zero. 


UNSYMMETRICAL   ARRANGEMENT.  821 

Hence  we  take  moments  about  the  inner  edge  of  each  of  the 
springing-joints. 

In  taking  moments  about  the  inner  edge  of  the  left-hand 
springing-joint,  we  impose  the  conditions  of  equilibrium  upon 
the  forces  acting  on  that  part  of  the  arch  that  lies  to  the  left  of 
the  crown-joint.  These  forces  are,  (i°)  its  load  and  weight, 
which  tend  to  cause  right-handed  rotation  ;  (2°)  the  horizontal 
component  of  the  thrust  exerted  by  the  right-hand  portion  upon 
the  left-hand  portion ;  (3°)  the  vertical  component  P  of  the  thrust 
exerted  by  the  right-hand  portion  upon  the  left-hand  portion. 

It  is  necessary  to  adopt  some  convention,  in  regard  to  the 
sign  of  P,  to  avoid  confusion  :  and  it  will  be  called  positive 
when  the  vertical  component  of  the  thrust  exerted  by  the  right- 
hand  portion  on  the  left-hand  portion  is  upwards ;  when  the 
reverse  is  the  case,  it  is  negative. 

We  next  take  moments  about  the  inner  edge  of  the  right- 
hand  springing-joint,  and  impose  the  conditions  of  equilibrium 
upon  the  forces  acting  upon  the  right-hand  portion  of  the  arch. 
In  doing  this,  we  must  observe  that  we  have  for  these  forces, 
(i°)  the  weight  and  load  which  tend  to  cause  left-handed  rota- 
tion ;  (2°)  the  horizontal  component  Q  of  the  thrust  exerted  by 
the  left-hand  portion  upon  the  right-hand  portion,  —  this  acts 
towards  the  right ;  (3°)  the  vertical  component  P  of  the  thrust 
exerted  by  the  left-hand  portion  upon  the  right-hand  portion ; 
and  this,  when  positive,  acts  downwards. 

Having  determined  the  values  of  Q  and  P,  we  next  proceed 
to  draw  the  line  of  resistance,  by  the  use  of  either  of  the  methods 
employed,  with  symmetrical  arches,  observing  only  that  the 
thrust,  i.e.,  the  resultant  of  Pand  Q,  is  now  oblique,  and  that  it 
acts  in  opposite  directions  on  the  two  sides  of  the  crown-joint. 

Having  drawn  this  line  of  resistance,  if  we  find  that  it  passes 
outside  of  the  arch-ring,  we  draw  normals  through  the  points 
where  it  is  farthest  from  the  arch-ring,  and  thus  obtain  three 
points  through  which  to  draw  a  line  of  resistance  :  then,  taking 


822 


APPLIED   MECHANICS. 


moments  about  each  of  these  three  points,  we  determine,  from 
the  three  resulting  equations,  values  of  <2,  Py  and  x,  and  pro- 
ceed to  draw  our  new  line  of  resistance ;  and,  if  this  does  not 
pass  entirely  within  the  arch-ring,  it  is  not  at  all  probable  that 
a  line  of  resistance  can  be  drawn  within  the  arch-ring.  All  the 
above  will  be  made  clearer  by  the  following  example :  — 

EXAMPLE.  —  Given  an  unsymmetrical  circular  arch,  shown 
in  the  figure,  the  intrados  and  extrados  being  concentric 
circles,  EM  =  4',  HF  =  i'.85,  radius  of  EHF  =  6,  AH  =  o'.s, 
AK  =  o'.S,  to  determine  the  possibility  of  drawing  a  line  of 


FIG.  290. 


resistance  in  the  arch-ring.  The  tables  following  show  the 
mode  of  dividing  up  the  load,  and  getting  the  centres  of 
gravity,  also  the  mode  of  arranging  the  work  for  this  pur- 
pose. 


UNS YMME  TRICA L   ARRA NGEMENT. 


823 


LEFT-HAND   PORTION. 


Number  of 
Voussoir. 

Width. 

Height. 

Area. 

Lever 

Arm. 

Moment. 

Partial 
Sums. 

Area. 

Moment. 

Lever 
Arm. 

I 

I.OO 

1.32 

1.32 

0.50 

0.660 

I 

1.32 

0.660 

0.50 

2 

I.OO 

!.48 

1.48 

1.50 

2.22O 

I   +  2 

2.80 

2.880 

1.03 

3 

I.OO 

1.84 

1.84 

2.50 

4.600 

1  +  2  +  3 

4.64 

7.480 

I.6l 

4 

I.OO 

2.42 

2.42 

3-50 

8.470 

i  +  ...  +  4 

7-06 

'5-950 

2.26 

5 

o-33 

2.63 

0.87 

4.17 

3.628 

I+...+  5 

7-93 

I9-578 

2-47 

- 

- 

- 

7-93 

- 

I9-578 

- 

- 

- 

- 

RIGHT-HAND   PORTION. 


"o    . 

Number 
Voussoi 

Width. 

Height. 

Area. 

Lever 
Arm. 

Moment. 

Partial 
Sums. 

Area. 

Moment. 

Lever 
Arm. 

*y 

I.OO 
I.OO 

1.32 

I.48 

1.32 
I.48 

0.50 
I.50 

0.660 
2.220 

I 

I    +  2 

1.32 
2.80 

O.66o 
2.880 

0.50 
1.03 

- 

- 

- 

2.80 

- 

2.880 

- 

- 

- 

- 

Now  take  moments   about   the   left-hand   springing   inner 
edge,  and  we  have 

2.02<2  +  4^=  7-93(4  -  2.47)  =  12.1239. 

Then  take  moments  about  the  right-hand  springing  inner  edge, 
and  we  have 

o.79<2  +  i.85/>=  2.8o(i.85  -  1.03)  =  2.3240. 

Solving  these  two  equations  gives  us 

Q  =  4.602, 
P  =  0.707. 


824  APPLIED   MECHANICS. 

If  R  represent  the  resultant  of  P  and  Q,  we  have 


R  =  \P*  +  <22  =  4-66 ; 

hence  we  proceed,  as  follows,  to  pass  a  line  ol  resistance 
through  the  top  of  the  crown-joint  and  the  inner  edge  of 
each  springing :  — 

Through  the  top  of  the  crown  draw  a  horizontal  line  BAG, 
Lay  off  Aa  —  4.602  and  ab  =  0.707,  and  draw  Ab ;  then 
Ab  =  4.66  represents,  in  direction  and  magnitude,  the  thrust 
at  the  crown.  Using  this  thrust  in  the  same  way  as  we  did  the 
horizontal  thrust  in  the  case  of  symmetrical  arches,  we  obtain 
the  line  of  resistance  EdAF,  which  is  farthest  outside  of  the 
arch  at  d;  hence,  drawing  a  normal  to  the  arch  from  d,  we  ob- 
tain c,  the  upper  edge  of  the  first  joint  from  the  crown.  Hence 
we  proceed  to  pass  a  new  line  of  resistance  through  E,  c,  and  F. 

To  do  this  we  must  assume  Q,  P,  and  x  all  unknown. 

i°.  Take  moments  about  E,  and  we  have 

(2.02  —  x)Q  +  4P  =  12.1239. 

2°.  Take  moments  about  F,  and  we  have 
(0.79  —  x)Q  —  i.86P  =  2.3240. 

3°.  Take  moments  about  c,  and  we  have 

(0.078  —  x)Q  +  P=  (1.32)  (0.50)  =  0.66. 
Solving  these  three  equations,  we  obtain 

Q=  4-91, 
P  =  0.64, 

x  =  0.074. 
Hence 


R  =  \IP*  +   Q2  = 


4.95. 


Hence,  if  we  lay  off  a  distance  0.078  below  A,  we  shall  have 
the  point  on  the  crown-joint  at  which  the  thrust  is  applied; 


GENERAL   REMARKS.  825 


and  making  the  same  kind  of  construction  as  we  just  made, 
only  using  this  point  instead  of  A,  and  these  new  values  of  Q 
and  P,  we  construct  the  second  line  of  resistance.  The  con- 
struction is  omitted  in  order  not  to  confuse  the  figure ;  but  the 
line  of  resistance  is  drawn,  and  the  student  can  easily  make 
the  construction  for  himself.  It  will  be  seen,  that,  in  this 
case,  this  new  line  of  resistance  lies  entirely  within  the  arch- 
ring. 

§  270.  General  Remarks.  —  Whenever  there  are  also  hor- 
izontal external  forces  acting  upon  the  arch,  these  should  be 
taken  into  account  in  imposing  the  conditions  of  equilibrium. 

It  will  be  noticed,  that,  in  the  preceding  discussion,  it  has 
always  been  assumed  that  the  load  upon  any  one  voussoir  is  the 
weight  of  the  material  directly  over  that  voussoir.  This  is  the 
assumption  usually  made  in  computing  bridge  arches  :  and  it 
may  be  nearly  true  when  the  height  of  the  load  above  the  crown 
is  not  great ;  but  even  then  it  is  not  strictly  true,  and  when 
this  depth  becomes  great,  as  would  be  the  case  with  an  arch 
which  supports  the  wall  of  a  building,  it  is  far  from  true,  as  the 
distribution  of  the  load  actually  coming  upon  different  parts  of 
the  arch  must  vary  with,  and  depend  upon,  the  bonding  of  the 
masonry,  and  also  upon  the  co-efficient  of  friction  of  the  mate- 
rial. Thus,  in  the  case  of  an  arch  supporting  a  part  of  the  wall 
of  a  building,  it  is  probable  that  the  only  part  of  the  load  that 
comes  upon  the  arch  is  a  small  triangular-shaped  piece  directly 
over  the  arch,  and  that  above  this  the  material  of  the  wall  is 
supported  independently  of  the  arch.  This  will  be  plain  when 
we  consider,  that,  were  such  an  arch  removed,  the  wall  would 
remain  standing,  only  a  few  of  the  bricks  near  the  arch  falling 
down ;  and  though  the  number  of  bricks  that  would  fall  would 
be  greater  while  the  mortar  is  green,  still  even  then  only  a  few 
would  drop  out. 

In  regard  to  these  matters,  we  need  experiments ;  but  thus 
far  we  have  none  that  are  reliable. 


826  APPLIED   MECHANICS. 

Then,  again,  we  have  arches  supporting  a  mass  of  sand  or 
gravel ;  and  then  the  mutual  friction  of  the  particles  on  each 
other  comes  into  play,  and  it  is  not  true  in  this  case  that  the 
load  on  any  voussoir  is  the  weight  of  the  material  directly 
above  that  voussoir.  In  some  cases  this  has  been  accounted 
as  a  mass  of  water  pressing  normally  upon  the  arch,  but  we 
canhot  assert  that  such  a  course  is  correct. 

On  the  other  hand,  there  are  cases  where  we  know  that. an 
arch  is  subjected  to  horizontal  as  well  as  to  vertical  forces,  and 
sometimes  we  cannot  tell  how  great  these  horizontal  forces  are. 
Thus,  the  forms  of  sewers  are  an  arch  for  the  top  and  an 
inverted  arch  for  the  bottom  ;  but  in  this  case  the  sides  of  the 
ditch  in  which  the  sewer  is  laid  when  building  it,  are  capable  of 
furnishing  whatever  horizontal  thrust  is  needed  to  force  the 
line  of  resistance  into  the  arch-ring,  provided  that  a  horizontal 
thrust  is  what  is  needed  to  force  it  in.  Hence  it  is,  that,  were 
the  attempt  made  to  pass  a  line  of  resistance  within  the  arch- 
ring  of  almost  any  successful  sewer,  accounting  the  load  as  the 
weight  of  the  earth  above  it,  the  line  would  almost  invariably 
go  outside ;  but  the  earth  on  the  sides  is  capable  of  furnishing 
the  necessary  horizontal  thrust  to  force  it  inside,  unless  a  care- 
less workman  has  omitted  to  ram  it  tight,  or  unless  some  other 
cause  has  loosened  it  on  the  sides  of  the  sewer. 

If  we  know,  in  any  case,  the  actual  law  of  the  distribution 
of  the  load,  we  can  determine  the  proper  form  for  the  arch  by 
the  methods  of  the  first  part  of  this  chapter,  as  was  done  in 
the  case  of  the  parabola  and  of  the  catenary.  Scheffler's 
method  is,  however,  the  one  almost,  always  used  for  determin- 
ing the  stability  of  any  stone  arch  against  overturning  around 
the  joints. 

Should  there  ever  arise  a  case  where  there  was  danger  that 
the  resultant  pressure  on  any  joint  made  an  angle  with  the 
joint  greater  than  the  angle  of  friction,  this  could  be  remedied 
by  merely  changing  the  inclination  of  the  joint. 


GENERAL    THEORY  OF   THE   ELASTIC  ARCH.  82/ 

§  271.  General  Theory  of  the  Elastic  Arch.  —  In  the  case 
of  the  iron  arch,  the  loads  upon  the  arch  are  all  definitely 
known  ;  and  it  is  necessary  to  ascertain  with  certainty  the  stress 
in  all  parts  of  the  structure,  and  to  so  proportion  the  different 
merribers  as  to  bear  with  safety  their  respective  stresses. 

The  general  discussion  of  the  method  used  in  calculating 
such  arches  will  now  be  given ;  the  method  used  being  practi- 
cally that  followed  by  Dr.  Jacob  J.  Weyrauch,  and  explained 
more  at  length  in  his  "Theorie  der  Elastigen  Bogentrager." 

This  discussion  is  also  necessary  in  order  to  prove  the 
proposition  already  enunciated  in  §  262  ;  viz.,  that  "for  an  arch 
of  constant  cross-section,  that  line  of  resistance  is  approximately 
the  true  one  which  lies  nearest  to  the  axis  of  the  arch-ring  as 
determined  by  the  method  of  least  squares." 

In  this  discussion  the  following  definitions  are  adopted  :  — 

i°.  The  axis  of  the  arch  is  a  plane  curved  line  passing 
through  the  centres  of  gravity  of  all  its  normal  sections. 

2°.  The  plane  of  this  axis  is  called  the  plane  of  the  arch. 

3°.  The  axial  layer  of  the  arch  is  a  cylindrical  surface  per- 
pendicular to  the  plane  of  the  arch,  and  containing  its  axis. 

4°.  A  section  normal  to  the  axis  is  called  a  cross-section. 

5°.  The  length  of  the  axis  between  two  sections  is  called 
the  length  of  arch  between  the  sections. 

The  loads  may  be  single  isolated  loads,  or  they  may  be 
distributed  loads. 

We  shall,  in  this  discussion,  assume  in  the  plane  of  the  arch 
a  pair  of  rectangular  axes,  OX  and  O  Y,  positive  to  the  right  and 
upwards  respectively. 

We  will,  then,  assuming  any  point  on  the  axis  of  the  arch 
before  the  loads  are  applied,  call  x,  y,  the  co-ordinates  of  that 
point,  s  the  length  of  axis  from  some  arbitrary  fixed  point,  <f>  the 
angle  made  by  the  tangent  line  at  that  point  with  OX,  r  the 
radius  of  curvature  of  the  axis  at  that  point,  x  -f-  dx,  y  -j-  dy, 
s  +  dsy  and  (/>  +  d$,  the  corresponding  quantities  for  a  point 


828 


APPLIED    MECHANICS. 


very  near  the  first  before  the  load  is  applied  ;  also  we  will  denote 

by  77  the  perpendicular  distance  of 
any  fibre  from  the  axial  layer,  by  Sr, 
the  length  of  arc  measured  to  that 
point  where  this  fibre  cuts  the  cross- 
section  through  (x,  y),  and  s^  + 
ds^  the  length  of  arc  measured  on 

this  fibre  to  the  next  cross-section, 

FlG  29X-  so  that  ds  will  be  the  distance  apart 

of  the  cross-sections  measured  on  the  axis,  and  ds^  on  the  other 
fibre.     All  this  is  done  before  the 
load   is  applied,  and    is    shown    in 
Fig.  291  ;  while  the  changes  brought 
about    by    the   application    of    the 
loads  combined  with  change  of  tem- 
perature are  denoted  by  ^'s,  and 
shown  in  Fig.  292.     Thus,  X,  y,  sy 
and    0   become    respectively  x  -\- 
Ax,  y  -f-  Ay,  s  -f-  As,  and  0  -f-  A<p. 
Now  the  course  we  are  to  fol- 
low in  the  discussion  is,  to  imagine  FlG- 292< 
a  cross-section  dividing  the  arch  into  two  parts,  and  to  impose 
the  conditions  of  equilibrium  between  the  external  forces  acting 
on  the  part  to  one  side  of  the  section,  and  the  forces  exerted 
by  the  other  part  upon  this  part  at  the  section.     These  latter 
forces  may  be  reduced  to  the  three  following :  — 

1°.  A  normal  thrust   Tx  uniformly  distributed  over  the  sec- 
tion, the  resultant  acting  at  the  centre  of  gravity  of  the  section. 
2°.  A  shearing-force  Sx  at  the  section 

3°.  A  bending-couple  at  the  section ;  this  comprising  a 
stress  varying  uniformly  from  the  axial  layer,  and  amounting 
to  a  statical  couple,  tension  below,  and  compressions  above,  the 
axial  layer. 

Moreover,  (i)  and  (3)  combined  amount  to  a  uniformly  vary- 


GENERAL    THEORY  Of    THE  ELASTIC  ARCH.  829 

ing  stress,  the  magnitude  of  whose  resultant  is  Txt  its  point  of 
application  not  being  at  the  centre  of  gravity  of  the  section  ; 
this  sort  of  composition  having  been  already  exhibited  in  the 
case  of  the  short  strut  (§  207). 

Now,  let  r  be  the  radius  of  curvature  of  the  axial  layer  at 
the  section  ;  and  we  have,  from  Fig.  291,  by  similar  sectors, 

ds^  =  ds  -f-  t)(  —  d<f>)  =  ds  —  r)d<}>.  (i) 

But 


Now,  if  the  loads  are  applied,  and  the  changes  take  place 
that  are  indicated  in  Fig.  292,  we  shall  have,  by  suitable  sub- 
stitutions in  (i), 

d(s,  +  AJ,)  =  d(s  +  AJ)  -  yitd  +  A<£)  ;  (3) 

and,  combining  this  with  (i)  and  (2),  we  obtain 

(4) 


d&s,        fd^s         d\4>\ 

—  =  I  --  —  r  --  1 


\  ds  ds  )r  +  V) 

Now,  the  change  of  length  of  fibre  from  ds,  to  d(s,  +  AJ,)  is 
clue  to  two  causes  :  (i)  the  change  of  temperature,  (2)  the  stress 
acting  on  the  fibre  normal  to  the  section. 

Let  e  =  co-efficient  of  expansion  per  degree  temperature. 
T  =  difference  of  temperature,  in  degrees. 
/,  =  intensity  of  stress  along  the  fibre  at  section. 
E  =  modulus  of  elasticity  of  the  material. 
Then 


_  A  = 


E         ds,         \  ds         '  ds  /r  +  rj 
Hence,  solving  for  /,,,  we  have 

T*-T£)-I-  +  jB<T'          (6) 

ds  ds  ]r  +  f] 


830  APPLIED   MECHANICS. 

this  being  the  expression  for  the  stress  per  square  inch  on  the 
fibre  whose  distance  is  rj  from  the  axial  layer. 

Hence  we  shall  have,  by  summation,  if  elementary  area  — 


f^Ad> 
ir 


and  for  the  moment  Mx  we  have,  by  taking  moments  about  the 
neutral  axis  of  the  section  (i.e.,  horizontal  line  through  its  cen- 
tre of  gravity), 


Let  ^dA  —  A,  r^  $•££*  =  Q,  and  observe  that  ^dA  —  o, 
r  +  TJ 

since  the  axis  passes  through  the  centre  of  gravity  of  the  sec- 
tion, and  we  have 

r*)dA 


4-  - 
r  r  r 


Making  these  substitutions,  we  have 

- 1-^)7  ~  (-jf  -  eT)A> 


Hence,  solving  for      £  and         ,  we  have 
ds  ds 

I  Mx\    i 

^+  +  *T=Y> 

Mx        cr 


GENERAL    THEORY  OF   THE   ELASTIC  ARCH.  83! 

,  from  Fig.  292,  we  have 

d(x  4-  A*)  =  d(s  +  AT)  cos  (<£  -f 
-f  Ay)  =  </(j  -f  A.r)sin  (</>  + 


but,  if  we  write  cos  A<£  =  i,  and  sin  A</>  =  A$, 


.    ,  .                                                            dy  dx 

sin (9  4"  A<p)   =  sin  <p  4-  Amcos©  =  —j  -\-  Ad)-j-« 

as  as 

Hence 


+ 
dfr 

+  —  ^  4-         - 
ds   "         \  ds 

or,  omitting  the  last  terms,  and  integrating, 

A*  =  -/A</>4>  4-  /K&,  (ir) 

Aj^  ==      f^dx+fYdy;  (12) 

and,  integrating  (9)  and  (10), 

A*   =  /Yds,  (13) 

A<£  =  fXds.  (14) 

In  these  four  equations  we  have 

A#  =  horizontal  deflection  due  to  the  loads, 

Ay  =  vertical  deflection  due  to  the  loads, 

AJ-   =  change  of  length  of  arc  due  to  the  loads, 

A</>  =  change  of  slope  due  to  the  loads. 

The  three  equations  which  we  shall  have  occasion  to  use  are 
{11),  (12),  and  (14),  and  if  we  make  the  integrations  between 
the  limits  x  and  o,  they  become,  by  changing  their  order, 


=  J00  4-       xds,  (15) 


832  APPLIED  MECHANICS. 

(16) 

(17) 


where  J00  is  the  change  of  slope  for  x  —  o. 
If,  now,  we  write 

_,        Mx         Mx          Tx 


P   =     M*  4-   Tx 

we  shall  have 

X=MI-  ~,  (20) 

Y=-Pl+er;  (21) 

or  if  we  neglect  the  effect  of  temperature,  we  may  write 

X  =  MIt  (22) 

Y=-P>.  (23) 

Moreover,  we   may  with   very  little   error  substitute  the 
moment  of  inertia  /  for  fl  in  the  value  of  MIt  i.e.,  writing  this 

_  Mx         Mx          Tx 


§  272.  Manner  of  using  the  Fundamental  Equations  to 
Determine  the  Stresses  in  an  Iron  Arch.  —  In  order  to  be 
able  to  determine  the  stresses  in  all  the  members  of  an  iron 
arch  with  any  given  loading,  we  need  to  determine  the  three 
quantities  Txt  5^,  and  Mx  for  each  section. 

Now,  if  we  let  Rx  represent  the  thrust  at  the  section,  we 
shall  have 


DETERMINA  TION  OF  STRESSES  IN  AN  IRON  ARCH.     833 


and,  if  we  let  Hx  and  Vx  represent  the  horizontal  and  vertical 
components  of  Rx  respectively,  we  have  that  we  need  to  deter- 
mine the  three  quantities  Hx,  Vx,  and  Mx  for  each  section. 

Let  us  suppose  the  arch  to  be  subjected  to  vertical  loads 
only,  and  let 

H  =  horizontal  component  of  thrust  at  all  points, 
V  =  vertical  component  of  left-hand  supporting  force, 
F,  —  vertical  component  of  right-hand  supporting  force, 
M  =  bending-moment  at  left-hand  support, 
M'  =  bending-moment  at  right-hand  support. 
Assume   origin    of   co-ordinates  at    left-hand   support,  and 
x  +  to  the  right,  and  y  -f-  upwards,  and  impose  the  conditions 
of  equilibrium  upon  the  forces  acting  on  the  part  of  the  arch 
between  the  section  and  the  left-hand  support ;  then  we  have, 
if   Wis  any  one  load,  and  a  the  x  of  its  point  of  application, 

Hx  =  H,  (2) 

X    —  «•!•        '  )  \3/ 

Hence  it  is  plain  that  the  three  quantities  which  we  need 
to  determine  are  H,  V,  and  M. 

Now  these  are  also  the  three  unknown  quantities  which  will, 
by  suitable  reductions,  become  the  three  unknown  constant 
quantities  in  equations  (n)  to  (14).  The  determination  of  these 
three  quantities  requires  three  conditions ;  what  these  condi- 
tions are  depends  upon  the  manner  of  building  the  arch,  as  will 
be  seen  from  the  following  three  special  cases :  — 

CASE  I.  —  Let  the  arch  be  jointed  at  three  points,  viz.,  the 
two  supports,  and  one  other  point  whose  co-ordinates  are  x  = 
Xi  and  y  =  yv  Then  we  know,  that,  for  all  points  where 
there  is  a  hinge,  there  can  be  no  bending-moment.  Hence 


834  APPLIED  MECHANICS. 

M  =  o,        M'  =  o,       and       MXi  =  o, 

which  are  the  three  required  conditions  ;  and,  if  these  be  im- 
posed, it  is  easy  to  obtain  Hx,  Vxt  and  Mx,  for  every  section. 

CASE  II. — Let  the  arch  be  jointed  only  at  the  ends.  Then 
M  =  M'  =  o  gives  us  two  conditions :  and  for  the  third  we 
have  J/  =  o  ;  i.e.,  if  we  put  /  for  x  in  equation  (16),  §  271, 
after  having  made  the  integrations,  we  have  the  third  equa- 
tion, as  this  expresses  simply  the  condition  that  the  sup- 
ports remain  at  the  same  horizontal  distance  apart  after  the 
load  is  put  on  as  before.  With  these  three  conditions  we  can 
determine  HM  Vxy  and  Mx  for  all  sections. 

CASE  III. — Let  the  arch  be  fixed  in  direction  at  the  ends. 
We  must  now  have  three  conditions.  These  will  be  as  follows: — 

i°.  Al  —  o  ;  i.e.,  the  supports  remain  at  the  same  horizontal 
distance  apart  after  the  load  is  applied  as  before. 

2°.  Ah  —  o  (h  being  the  difference  of  level  of  the  sup- 
ports) ;  i.e.,  the  supports  remain  at  the  same  vertical  distance 
apart  after  as  before  the  load  is  applied. 

3°.  J0,  =  o;  i.e.,  the  tangents  at  the  ends  make  the  same 
angle  with  each  other  after  as  before  the  load  is  applied. 

The  value  of  A^  is  obtained  by  integrating  (15),  §  271,  and 
then  substituting  /  for  x,  or  h  for  y,  observing  that  ^/00  —  o. 

The  value  of  Al  is  obtained  by  integrating  (16),  §  271,  and 
then  substituting  /  for  x. 

The  value  of  Ah  is  obtained  by  integrating  (17),  §  271,  and 
then  substituting  /  for  x,  or  h  for  y. 

In  this  case,  if  we  neglect  the  effect  of  temperature,  write 

/  for  .Q,  omit  all  terms  containing  -,    and  also  neglect   Tx 


r  -  - -  ! 


(15),  (16),  and  (17)  of  §  271,  we  shall  obtain  by  making  one 
integration, 


DETERMINATION   OF  STRESSES  IN  AN  IRON  ARCH.    835 


=+/§*<*> 


<6> 


While  it  has  often  been  proposed  to  use  these  as  approxi- 
mately true,  nevertheless  the  degree  of  approximation  is  too 
coarse  to  render  them  suitable  to  use  in  practice. 

CIRCULAR    ARCH,    UNIFORM    SECTION,    AND   VERTICAL   LOADS. 


We  will  next  deduce  expressions  for  J0,  Ax,  and  Ay,  for  a 
circular  arch  of  constant  cross-section  and  loaded  vertically, 
and  thence  deduce  the  equations  from  which  to  determine  the 
three  quantities  Mt  M' ,  and  H  in  any  such  case,  and  also  the 
expression  for  the  horizontal  thrust  in  an  arch  hinged  at  the 
two  springing-points,  and  symmetrical  in  form.  We  will  write 
in  place  of  £1  the  moment  of  inertia  /,  and  will  neglect  terms 

containing -,  in   equations   (15),  (16),  and   (17),  but   will   not 

'neglect  Tx. 

Take  the  origin  at  the  left-hand  springing-point,  and  the 
axis  of  x  horizontal. 

Observe  that  if  0  represent  the  angle  the  tangent  line  to 
the  arch  at  the  point  (x,  y)  makes  with  the  axis  of  x,  it  also 
represents  the  angle  subtended  by  the  radius  drawn  through 
the  point  (x,  y)  with  the  vertical  radius,  i.e.,  that  through  the 
crown. 

Let  00  be  the  value  of  0  at  the  origin,  and  let  a  be  the 
value  of  0  at  the  point  of  application  of  any  concentrated  load 
W,  the  co-ordinates  of  this  point  being  (a,  b). 


836  APPLIED  MECHANICS. 

Let  the  co-ordinates  of  the  centre  of  the  circle  be 

g  =  r  sin  00,  —  k  =  —  r  cos  00; 

of  the  crown  be        g  =  r  sin  00,  f=r  —  r  cos  00; 

of  the  point  of  ap- 

plication of  Wltz  a  —  r(sin  00  —  sin  a),     b  =  r(cos  a  —  cos  00); 
and   of    any   point 

on  arch,  x  =  r(sin  00  —  sin  0),    y  =  r(cos  0  —  cos  00). 

The  following  is  a  list  of   relations  which  can  be  easily 
proved,  and  which  are  needed  for  use  in  the  work  that  follows 
them. 
g  —  x  =  r  sin  0;  Tx  =  Vx  sin  0  +  H  cos  0; 

£  4"  V   —  r  COS   0:  ,-.  ,  ,    ,       Tr  rr  -Z?  rrr/  \ 

,.  .      o.      J/^  =  Jf+  F*  —  JTy—  ^^(tf  —  a);, 

x  —  a  —  r(sm  a  —  sin  0)  ;  .  0 

£--*  =  rsin*;  ^  =  F 

y  —  b  =•  r(cos  0  —  cos  «)  ; 


dx  =•  —  r  cos 

^  =  —  r  sin  0</0; 

where  M*  —  bending-moment  at  right-hand  springing-point. 
Also 


By  making  the  substitutions  indicated,  and  also  the  inte- 
grations, we  obtain  from  (15),  (16),  and  (17)  the  following:  — 


-      {  (00  -  0)(J/+  Fr  sin  00  +  Hr  cos  00) 
—  Vr  (cos  0  —  cos  00)  —  Zfr  (siA,00  —  sin  0)  +  2  Wr  (cos  0 

o 

-  cos  of)  -  :kwr  (a  -  0)  sin  a}9  (8) 

o 

Ax  =  erx  -  -^  {  F(sina  00  -  sin9  0)  +  J7[sin  00  cos  00 

—  sin0cos0-f  (00—  0)]—  J"^(sin2af  —  sin20)}  — 
sin  00  +  Hr  cos  00)  [(00  -  0)  cos  0 


DETERMINATION  OF  STRESSES  IN  AN  IRON  ARCH.    837 

-  (Sin00-sin0)]-^(cos0-  0J'  -^  [-  (00  -  0) 
+  cos  0  (sin  00  —  sin  0)  +  sin  00  (cos  0  —  cos  00)] 
-f  %r2  W  (cos  0  —  cos  a)*  —  r  cos  ^W  (a  —  0)  sin  a 

o  o 

+  r2  W  sin  a  (sin  a  —  sin  0) } ;  (9) 

Ay  =  ery  -  -        j^T(sina  00  -  sin2  0)  -  F  [sin  00  cos  00 


—  sin  0  cos  0  —  (00  —  0)]  —  2  W  [sin  0  cos  0  +  sin  a  cos  a 

0 

+  (<*  —  0)] }  +  *^00+  1F>  K^+  ^  sin  00 


cos  00)  [(cos  0  —  cos  00)  —  (00  —  0)  sin  0] 


Vr 
—  \Hr  (sin  00  —  sin  0)*  --  ~~\.~~  cos 


—  sin  0  (cos  0  —  cos  00)  -f  (00  —0)1  +  sin  ^Wr  (a 

o 
je 

—  0)  sin  a  —  2  Wr  sin  a  (cos  0  —  cos  a) 

o 
* 

—  ^2  Wr  [cos  «  (sin  a  —  sin  0)  +  sin  0  (cos  0  —  cos  a) 

°  (10) 


We  will  next  write  out  the  same  values  as  applied  to  the 
right-hand  springing-joint  of  a  symmetrical  arch. 

In  this  case  we  have  the  value  of  0  for  the  right-hand  end, 
or  0,  equal  to  —  00;  and  if  we  make  this  substitution,  observing 
that  x  becomes  /and  y  becomes  zero,  and  if  we  substitute  for 
Fthe  value 


v_     M*  -  M       i^Wr  (sin  ^o  +  sin  a) 
r  sin  00  r  sin  00 

then  we  obtain  the  following  :  — 

-\(Mf  +  M)  00  -  2Hr  (sin  00  -  00  cos  00) 


sin  or  —  200  sin  00  -f-  2  (cos  «  —  cos  00)};  (n) 


838  APPLIED  MECHANICS. 


=  erl- 

-f  -—  {  (M'  +  M)  (2  sin  0o-200  cos  00)  -  Hr(^Q  cos2  0 


-  6  sin  00  cos  00  +  200)  +  2Wr  \_2a  cos  00  sin  a 

o 

+  2  cos  00  cos  or  —  20o  sin  00  cos  00  +  sin2  00  —  sin2  a 
-2cos800](;  (12) 

0o  sin  00)  -  4^>  (sin2  00 


-  00  sin  00  cos  00)  -  (Jf  -  M9)  (cos  00  -  - 

-  12  Wr  [200  (-^-^  -  2  sin9  00)  +  2*  (2  sin  00  sin  a  -  i) 

o  \sin  (pQ  / 

—  4  sin  00  cos  00  +  2  sin  a  cos  00  —  2  sin  <*  cos  a 

+  4  cos  a  sin  0J}  -  ^  {(Jf-  ^)  (cos  00  -  ^-) 

-^-  sin  «  —  20?  —  2  sin  a  (cos  or  —  cos  00)]    .  (13) 


SPECIAL    CASES    OF    SYMMETRICAL    ARCHES. 

i°.  Three-hinged  Arch.  —  In  this  case  we  do  not  need  these 
equations  to  find  the  horizontal  thrust  :  the  proper  ones  can 
be  used  subsequently  if  we  wish  the  deflections  or  slopes. 

2°.  Arch  hinged  at  the  two  springing-points.  —  In  this  case 
M  —  M'  —  o  ;  and  by  making  these  substitutions  in  (12)  and 
solving  for  Ht  we  obtain 

2W[2<x  cos  00  sin  a-\-2  cos  00  cos  a—  200  sin  00  cos  00 
+  sina  00  —  sin8  a  —  2  cos8  00] 

—  ^2W(sm*  00  -  sin2  a)  -  —  (J/  -  erl) 
ff=-  --  -(14) 


400cos800  —  6  sin00cos00+200+-i  (200+2  sin00cos00) 


DETERMINATION  OF  STRESSES  IN  AN  IRON  ARCH.    839 

This  formula  gives  the  thrust  when  the  value  of  Al  is  known, 
i.e.,  the  amount  of  relative  yielding  of  the  supporting  points. 

If  the  abutments  do  not  yield  at  all,  then  Al  =  o,  and  that 
term  should  be  omitted  from  the  numerator  ;  so,  also,  if  we 
neglect  a  consideration  of  the  temperature,  then  erl  vanishes 
in  addition. 

Formula  (14)  gives  the  thrust  for  a  set  of  concentrated  loads, 
each  equal  to  W. 

For  a  distributed  load,  we  should  substitute  for  Wy  wdx, 
and  integrate  between  the  proper  limits,  w  being  the  intensity 
of  the  load  per  unit  of  horizontal  length,  and  being  constant  or 
variable  according  to  the  distribution  of  the  load. 

The  formula  .for  the  thrust  in  the  case  where  the  load  is 
uniformly  distributed  horizontally  (i.e.,  when  w  is  a  constant) 
and  when  it  covers  the  entire  arch  will  now  be  given,  but  will 
not  be  worked  out  here,  as  it  is  easily  obtained  from  (14). 

In  this  formula  the  letters  have  the  same    meanings   as 

heretofore,  and  we  use  also  Al  =  I  ydx  =  area  of  segment  of 

t/o 

arch. 

The  formula  is  as  follows  :  — 


sin  200-00  (/•+,*  cos  200) 


m 


In  a  similar  way  formulae  are  easily  obtained  for  the  thrust 
when  half  or  a  quarter,  or  some  other  portion  of  the  arch,  is 
loaded. 

3°.  Arch  with  no  hinges.  —  In  this  case,  if  we  know  A(p,  Al, 
and  Ac,  or  if  these  are  zero,  we  can  obtain  from  (n),  (12),  and 
(13)  the  three  quantities  M,  M',  and  H,  and  then  the  solution  of 
the  arch  follows. 

This  will  not  be  done  here,  however,  as  the  arches  usually 
built  are  of  the  other  kinds. 


840  APPLIED  MECHANICS. 

EXAMPLES. 

i.  Given  a  semicircular  arch  jointed  at  each  springing- joint  and  at 
the  crown,  radius  r.  Trace  out  the  effect  of  a  single  load  W  acting 
upon  it  at  the  extremity  of  a  radius  making  45°  with  the  horizontal. 

Solution. 

The  presence  of  three  joints  gives  us  the  bending-moments  at  each 
of  these  joints  equal  to  zero,  the  co-ordinates  of  these  joints  being 
respectively  (o,  o),  (r,  r),  and  (27-,  o). 

Hence,  using  equation  (4),  we  obtain 

i°.   M  =  o, 

2°.    Vr-  Hr  -   ^(0.7071  ir)  =  o,  <  ).* 

3o.    V(2r)  -  ^(1.7071  ir)  =  o. 

Solving,  we  have,  therefore, 

V  =  0.85  355  W=  left-hand  supporting-force, 
and 

H  —  0.14645  W  =  horizontal  component  of  thrust. 

Hence   Fx  =  0.14645  W=  right-hand  supporting-force. 
Hence,  for  a  section  whose  co-ordinates  are  (x,y), 

x  <  0.29289?-,          Vx  =       0.85355  W; 
x>  0.29289^-,          ^  =  —0.14645^. 

Hence  equation  (i)  gives,  for 


x  <  0.292897-,        Rx  =  £FV(o.85355)2  +  (0.14645)* 

=  0.86602  W, 
x  >  0.29289;-,         Rx  =  0^(0.14645)'  +  (o.i4645)2 

=  0.20711  W. 
Now,  the  angle  made  by  Rx  with  the  horizontal  is,  for 


*<0.29289,,          a,  =  tan-  =  80°  15'  51", 


TRUE  LINE   OF  RESISTANCE  IN  A   STONE  ARCH.       841 

Knowing,  now,  the  angle  made  by  Rx  with  the  horizontal,  we  can 
find,  for  the  point  (x,  y),  the  angle  made  by  a  tangent  to  the  circle  with 

the  horizon,  or  a2  =  tan-'f J.  Then  resolve  Rx  into  two  com- 
ponents, respectively  tangent  to  the  arch  and  normal  to  it  at  the  point 
x,  y,  and  the  tangential  component  is  the  direct  thrust  Txt  while  the 
normal  is  the  shearing-force  Sx. 

Then,  for  the  bending  moment,  we  have,  from  (4), 

x  <  0.29289?-,   Mx  =  0.85355  Wx  —  0.14645  Wy; 

,x  >  0.29289?-,   Mx  —  0.85355  Wx  —  0.14645  Wy  —  W(x  —  0.29289?-). 

Hence  we  determine  the  direct  thrust,  the  shearing-force,  and  the 
bending-moment  at  any  section,  and  can  hence  obtain  the  stresses  at  all 
points. 

2.  Given  the  same  arch  with  a  load  W  distributed  uniformly  over 
the  circular  arc,  find  stresses  at  all  points. 

3.  Given  the  same  arch  jointed  only  at  the  two  springing-points, 
find  stresses  at  all  points. 

§  273.  Position  of  True  Line  of  Resistance  in  a  Stone 
Arch.  —  The  proof  will  now  be  given  of  the  proposition  already 
referred  to  in  regard  to  the  position  of  the  true  line  of  resist- 
ance ;  viz.,  —  I 

"  For  an  arch  of  constant  section,  that  line  of  resistance  is 
approximately  the  true  one  which  lies  nearest  to  the  axis  of  the 
arch-ring,  as  determined  by  the  method  of  least  squares." 

*PROOF.  —  If  we  denote  by  y  the  ordinate  of  the  axis  of  the 
arch  for  an  abscissa  x,  and  by  /x,  that  of  the  line  of  resistance 
for  the  same  abscissa,  then  /a  —  y  is  the  vertical  distance  be- 
tween the  two  curves  for  abscissa  x.  Now,  the  condition  that 
the  line  of  resistance  should  be  as  near  the  arch-ring  as  possi- 
ble, is,  that  the  sum  of  the  (/*  —  y)2  sna^  be  a  minimum,  or 

/(/A  —  y)2ds  =  minimum.  (i) 

But  (x,  p)  are  the  co-ordinates  of  the  point  of  application  of  the 


842  APPLIED   MECHANICS. 

actual  thrust,  and  hence  (/*  —  y)  is  the  distance  of  the  point  at. 
which  the  resultant  thrust  acts  from  the  centre  of  gravity  of 
the  section.  Hence  we  have 

M* 

»-y  =  -H' 
Hence  (i)  becomes 

CIMX\*  .  .  ,  x 

I  f  —  -  j  ds  =  minimum.  (2) 

But  H  is  constant  for  the  same  line  of  resistance,  though  it 
varies  for  different  lines  :  hence  we  can  place  H  outside  of  the 
integral  sign.  Hence  we  may  write 

u  =  —  I  Mx2ds  =  minimum.  (3) 

HZJ 

Now,  from  (4),  §  272,  we  have 

Mx=M+Vx-Hy  —  -$*W(x  -  a)  =  <j>(M,  F,  J3); 

M,  V,  and  H  being  constants  for  the  same  line  of  resistance,, 
but  varying  for  different  lines.     Hence,  by  differentiating  (3), 
we  have 
du          du  dMx 


du        du  dMx  . 

=  o     (5) 


-2H~*  IMJ4*  - 


du         du    dMx 
~dH~  ~dMx~dH 


But  the  first  term  must  be  very  small  :  hence  we  may  write  ap- 
proximately, 

fMxyds  =  o.  (6) 

Now,  the  three  expressions  (4),  (5),  and  (6)  are  identical  with 
(5)»  (6)»  and  (7)  of  §  272  ;  and  the  conditions  that  these  shall 
be  zero  are,  with  the  degree  of  approximation  there  stated,  the 


DOMES. 


843 


Sup- 


conditions  that  hold  in  the  case  of  an  arch  fixed  in  direction  at 
the  ends.  Hence  it  follows  that  the  condition  that  the  line  of 
resistance  shall  fall  as  near  the  centre  of  the  arch  as  possible  is 
the  condition  which,  in  an  elastic  arch  fixed  in  direction  at  the 
ends,  gives  us  its  true  position.  Hence  it  would  seem  that 
the  most  probable  position  for  the  true  line  of  resistance  is  the 
nearest  possible  to  the  axis  of  the  arch. 

This  is  the  conclusion  reached  by  Winkler ;  and  a  more 
detailed  discussion  of  the  matter  is  to  be  found  in  an  article 
by  Professor  Swain  in  "Van  Nostrand's  "  for  October,  1880. 

§  274.  Domes.  —  The  method  to  be  used  for  determining 
the  stability  of  a  dome  differs  essentially  from  that  used  in  the 
case  of  an  arch,  for  there  is  no  thrust  at  the  crown  in  a  dome. 
Indeed,  the  most  general  case  is  that  of  the  dome  open  at  the 
top.  we  will,  therefore,  consider  this  case  first  in  studying. the 
action  of  the  forces  required  to  preserve  equilibrium. 

Fig.  293  shows  a  meridional  section  of  an  open  dome, 
pose  that  this  dome  had  been 
entirely  built,  except  the  up- 
per ring-course  of  stones,  rep- 
resented by  LKGH.  Then, 
suppose  that  one  of  the  stones 
only  of  this  course  were  placed 
in  position  without  any  auxil- 
iary support,  its  own  weight 
would  evidently  overturn  it, 
since  the  line  ab,  along  which 
the  weight  acts,  does  not  cut 
the  joint ;  but,  if  the  whole 
ring-course  is  put  in  place,  the 
stones  keep  each  other  in  po-M- 
sition.  The  way  in  which  this  FIG 

is  accomplished  is  as  follows : 
they  press  laterally  against  each  other;  and  the  resultant  of  the 


#44  APPLIED  MECHANICS. 

pressures  exerted  upon  the  two  lateral  faces  of  any  one  stone 
by  the  other  stones  of  the  course  is  a  horizontal  radial  force, 
which,  combined  with  the  weight  of  the  stone,  gives,  as  the 
•resultant  of  the  two,  a  force  which  cuts  the  joint  between  G 
and  H.  Moreover,  sufficient  pressure  will  be  developed  to 
accomplish  this  result,  as  a  failure  to  reach  the  result  will  only 
increase  the  pressure  upon  the  lateral  faces. 

Moreover,  if,  when  sufficient  pressure  has  been  developed 
to  bring  the  resultant  of  the  weight  of  the  stone  and  the  above- 
described  horizontal  radial  force  within  the  joint,  it  should 
make  an  angle  with  the  normal  to  the  joint  greater  than  the 
angle  of  friction,  the  tendency  of  the  stone  to  slide  will  increase 
the  lateral  pressure,  and  this  in  turn  will  increase  the  outward 
horizontal  force  till  the  angle  made  by  the  resultant  with  the 
normal  to  the  joint  is  no  greater  than  the  angle  of  friction  of 
the  material  of  the  voussoirs. 

This  will  be  made  plain  by  reference  to  the  figure  (Fig. 
243),  where  ab  represents  the  weight  of  the  stone  HLKG,  and 
where  O/3  is  perpendicular  to  HG  and  Oy  is  drawn  so  that  -yOfi 
=  </>,  the  angle  of  friction.  Now,  since  ab  produced  passes  out- 
side of  HG,  horizontal  thrust  must  be  developed.  And,  more- 
over, were  only  sufficient  horizontal  thrust  furnished  to  make 
the  resultant  cut  HG  at  G,  the  angle  between  this  resultant 
-and  the  normal  to  the  joint  would  be  greater  than  <£;  there- 
fore we  proceed  as  follows :  assuming  the  horizontal  thrust  to 
act  through  L,  the  upper  edge  of  the  stone,  we  lay  off  from  b, 
the  intersection  of  the  horizontal  through  L  with  a  vertical  line 
drawn  through  the  centre  of  gravity  of  the  stone,  the  weight  ab 
to  scale,  then  from  b  draw  be  parallel  to  Oy,  and  draw  through 
a  a  horizontal  line  to  meet  be.  Then  will  ac  be  the  horizontal 
force  that  will  be  furnished  by  the  other  stones  of  the  course 
to  keep  this  stone  in  place ;  and  the  pressure  upon  joint  HG 
is  be,  and  acts  at  the  intersection  of  be  and  HG. 

Now   prolong  be  to  meet  the  vertical  drawn  through  the 


DOMES.  845 


centre  of  gravity  of  the  next  stone,  HGFE,  at  d.  Combine  it 
with  the  weight  of  this  stone ;  this  is  done  by  laying  off  de  =.  be, 
and  from  e  drawing  ef  vertical,  and  equal  to  the  weight  of 
FHGE.  The  resultant  fd  makes  an  angle  with  the  normal  to 
FE  greater  than  <£ :  hence  draw  O8  perpendicular  to  FE  and 
O^,  so  that  €08  =  </>;  then  from^-,  the  intersection  of  ^with  a 
horizontal  line  through  H,  the  top  of  FHGE,  lay  off  gs  =  df, 
through  g  draw  gh  parallel  to  <9«,  and  through  s  draw  sh  hori- 
zontal. Then  is  sh  the  horizontal  thrust  that  will  be  furnished 
at  H  to  keep  the  stone  HGEF  in  place ;  and  this  is  the  pres- 
sure upon  joint  FE,  and  acts  at  the  intersection  of  FE  with  kg. 

Next,  prolong  hg  to  meet  the  vertical  through  the  centre  of 
gravity  of  stone  FEDC  at  k ;  lay  off  kl  =  gh,  and  from  /  lay 
off  Im  ==  weight  of  stone  FEDC ;  draw  km,  which  cuts  the 
joint  within  the  joint  itself,  and  needs  no  horizontal  thrust  to 
bring  it  inside  ;  hence  mk  is  the  pressure  on  joint  DC. 

Then  draw  mk  to  meet  the  vertical  through  the  centre  of 
gravity  of  ABCD  at  n,  and  lay  off  no  =  km;  draw  op  =  weight 
of  ABCD,  and  draw  pn,  which  will  be  the  pressure  on  the  joint 
BA. 

It  is  necessary,  for  stability,  that  all  these  forces  should  cut 
the  joint  inside  of  the  joint  if  the  stones  are  reckoned  incom- 
pressible ;  or  we  may  adopt  the  middle  third,  or  other  limits,  as 
our  criterion  of  stability. 

As  long  as  it  is  outward  thrust  that  is  required  to  produce 
stability,  it  is  possible  to  furnish  it ;  but,  if  we  should  reach  a 
joint  where  inward  thrust  would  be  required,  this  could  not  be 
furnished,  and  the  dome  would  be  unstable.  Moreover,  the 
resultant  pressure  on  the  springing  gives  us  the  pressure  ex- 
erted upon  the  support  of  the  dome ;  and  it  must  not  cut  any 
joint  of  the  support  outside  of  that  joint,  as  otherwise  the  sup- 
port would  not  stand. 

In  determining  the  numerical  value  and  direction  of  this, 
pressure  on  the  support,  we  may  either  construct  it  graphically,, 


846  APPLIED   MECHANICS. 

or  we  may  compute  it  as  follows :  (i°)  Compound  all  the  ver- 
tical forces,  i.e.,  the  weights,  and  find  the  magnitude  and  line 
of  action  of  the  resultant  of  these.  (2°)  Compound  all  the 
horizontal  forces,  and  find  the  magnitude  and  line  of  action  of 
their  resultant  (in  this  case  the  horizontal  forces  are  two ;  viz., 
ac  applied  at  L,  and  sh  applied  at  H} ;  then  compound  these  two 
resultants.  The  graphical  and  analytical  method  should  check 
if  no  mistake  has  been  made  in  the  work. 

In  the  above  calculation,  it  has  been  assumed  that  the  figure 
represents  the  portion  of  a  dome  included  between  two  merid- 
ional planes. 

If  we  desire  to  ascertain  the  pressure  exerted  upon  the 
lateral  face  of  the  stone  by  its  neighbors  in  the  same  ring- 
course,  we  only  need  to  know  the  angle  made  by  the  two 
meridional  planes  containing  the  lateral  faces  of  the  stone  in 
question,  then  resolve  the  horizontal  thrust  upon  that  stone 
into  two  equal  components,  which  make  with  each  other  an 
angle  equal  to  the  supplement  of  the  angle  of  the  planes;  i.e., 
resolve  the  outward  horizontal  thrust  into  two  components 
normal  to  the  lateral  faces. 

In  regard  to  the  assumption  that  the  outward  thrust  acts  at 
the  top  of  the  stone,  it  should  be  said  that  this  is  SchefHer's 
custom,  his  reason  being  that  less  thrust  will  be  required  if  he 
assumes  it  at  the  top  than  if  he  assumes  it  nearer  the  middle. 
The  true  position  of  this  thrust  is  probably  much  nearer  the 
middle  of  the  stone. 

An  example  will  next  be  solved,  giving  Scheffler's  method 
of  working. 

EXAMPLE.  —  Given  the  dome  shown  in  the  figure,  sur- 
mounted by  a  lantern  at  the  top ;  determine  whether  it  is 
stable,  and  what  should  be  the  thickness  of  the  support  in  order 
that  the  resultant  pressure  may  not  pass  outside  any  joint  of 
the  pier. 

The  dimensions  are  as  follows  :  — 


DOMES. 


847 


Diameter  of  outer  vertical  circle  =  20  feet. 

Diameter  of  inner  vertical  circle  =  18  feet. 

Angle    made   by  springing-radius   with  vertical  =  75°  = 
angle  A  OB. 

The  inner  edge  of  the  upper 
voussoir  subtends  18°  on  the 
lower  circle  ;  the  width  of  the 
load  of  the  lantern  is  0.6 ; 
the.  voussoirs  below  that,  each 
subtend  18°. 

Assume  36  stones  in  a  hori- 
zontal course.  The  width  of  the 
lowest  will,  then,  be  1.51  ;  the 
width  of  the  others  are  deter- 
mined from  their  lever  arms. 

Given  height  of  pier  =  8 
feet. 

Height  of  the  centre  of  the 
sphere  above  base  of  pier  =  8' 
-  10  sin  15°  =  5.41'. 

The  figure  may  be  taken  to 
represent  the  portion  of  the 
dome  included  between  two  ver- 
tical planes  passing  through  the 
axis  of  the  dome  :  hence  it 
shows  one  vertical  series  of 
stones. 

We   first   construct  a  table 
giving  the  weights  of  the  differ- 
ent voussoirs  with  any  superin- 
cumbent load,  their  centres  of  gravity,  and  the  moments  of 
their  weights  about   an  axis   passing  through  O,  and   perpen- 
dicular to  the  central  plane  of  the   portion  shown  ;  and   we 
so    choose    our   unit    of    weight   that    the   volumes    of    the 


FIG.  294. 


848 


APPLIED  MECHANICS. 


voussoirs  shall  represent  their  weights, 
as  follows  :  — 


The  work  is  arranged 


ELEMENTARY  FORCES. 

HORIZONTAL  FORCES. 

(1) 

(2) 

(3) 

(4) 

(6) 

(6) 

(7) 

(8) 

(9) 

Voussoir. 

Area  of 
Lateral  Face. 

Thick- 
ness. 

Product. 

Lever 
Arms. 

Moment. 

Hori- 
zontal 
Forces. 

Lever 

Arms. 

Moment. 

I 
2 

3 

4 

0.6X6.680 
2.985 
2.985 
2.985 

°-53 
0.74 
1.23 
i-S1 

2.124 
2.209 
3.672 
4-5°7 

3-07 
475 
7-05 

8.68 

6.521 
10.493 
25.888 
39.121 

1.74 
1.26 
1.32 

9.60 

9-33 

7.78 

16.104 

11.756 
10.273 

- 

-    « 

- 

12.512 

- 

82.023 

4-32 

- 

38.730 

Column  (i)  contains  the  numbers  of  the  voussoirs,  counting 
from  the  top. 

Column  (2)  contains  the  areas  of  the  lateral  faces  of  the 
stones  shown  in  the  figure.  For  the  three  lower  stones,  the 
area  of  a  ring  subtending  18°  at  the  centre,  and  of  the  dimen- 
sions given,  is  calculated.  For  the  first,  the  height  is  6.68  and 
the  width  0.6. 

Column  (3)  contains  the  thicknesses  of  the  voussoirs ;  i.e., 
the  length  of  arc  between  their  two  lateral  faces  measured  on  a 
horizontal  circle  through  the  centre  of  gravity  of  the  voussoir, 
which  is  here  taken  at  the  middle  point  of  the  arc  subtended 
by  this  voussoir  on  its  middle  vertical  circle,  i.e.,  one  which 
has  a  radius  9.5  feet. 

Hence,  the  thickness  of  the  lower  stone  being  1.51  feet,  that 
of  the  others  will  be 


DOMES. 


849 


Column  (4)  gives  the  weights  of  the  voussoirs  and  their 
loads  :  it  is  obtained  by  multiplying  together  the  numbers  in 
columns  (2)  and  (3). 

Column  (5}  gives  the  distances  of  the  centres  of  gravity  of 
the  different  voussoirs  from  the  axis  of  the  dome :  it  may  be 
determined  graphically  or  by  calculation. 

Column  (6)  gives  the  moments  of  the  weights  about  a  hori- 
zontal axis  through  O  perpendicular  to  the  central  plane  of  this 
series  of  voussoirs.  The  graphical  construction  for  determin- 
ing the  horizontal  thrusts  required  is  next  made,  and  the  results 
are  recorded  in  column  (7).  It  will  be  seen  that  no  thrust  is 
required  on  voussoir  No.  4. 

Column  (8)  contains  the  lever  arms  of  these  forces  about 
the  same  axis. 

Column  (9)  contains  their  moments  about  the  same  axis. 

The  construction  thus  far  has  shown  no  case  where  horizon- 
tal tension  instead  of  horizontal  thrust  is  required  to  cause  the 
thrust  on  any  joint  to  pass  within  the  joint :  hence  thus  far  the 
dome  is  stable ;  and  the  question  comes  next  as  to  what  should 
be  the  width  of  the  pier  in  order  that  the  line  of  resistance,  if 
continued  down,  may  remain  within  it. 

For  this  purpose  we  proceed  as  follows :  — 

Let  t  —  thickness  required. 

Let  breadth  be  equal  to  that  of  the  lowest  voussoir. 

Height  =  8  feet. 

Take  moments  about  the  outer  edge  of  the  base  of  the  pier. 

We  shall  then  have,  — 

i°.  Moment  of  vertical  load  on  dome,  and  of  weight  of  dome 
sector  about  inner  edge  of  springing,  — 

(12.512)  (8. 69  —  6.56)  =  26.52. 
2°.  Moment  of  same  about  outer  edge  of  springing  of  pier  = 

26.52    +    (I2.5I2)/. 


850  APPLIED   MECHANICS, 

3°.  Moment  of  horizontal  forces  about  the  same  axis  =. 

38.730  +  (4.32)  (5.41)  =  62.101. 
4°.  Moment  of  weight  of  pier  about  outer  edge  = 

=  6.04*. 


Hence  we  have 

6.04/2  -f-  i2.5i/'-h  26.52  =  62.191 

.-.     t*  +  2.07/  =  5.89  .*.     /  =  i.  60  feet. 

This  is  the  thickness  required  in  order  that  the  line  of 
resistance  may  remain  within  the  lower  joint. 

If,  on  the  other  hand,  while  pursuing  the  same  method  with 
the  dome  itself,  we  require  that  the  line  of  resistance  shall 
remain  within  the  middle  third  of  the  pier,  we  take  moments 
about  a  point  in  the  springing  of  the  pier  at  a  distance  f  /  from 
its  inner  edge,  we  should  then  have 

f/*    +   f(2.07)/  =    5.89       • 

/.     /2  +  2.07/  =  8.84  .-.     t  =  2.10  feet. 

On  the  other  hand,  we  could  proceed  in  a  similar  way  to 
the  above,  if  we  desired  to  keep  the  line  of  resistance  in  the 
dome  within  the  middle  third,  by  merely  assuming  the  horizon- 
tal thrusts  to  act  at  two-thirds  the  thickness  of  a  joint  from  the 
lower  edge,  and  using  a  point  two-thirds  the  thickness  from 
the  top,  instead  of  the  lower  edge,  as  the  lower  limiting-point 
for  the  pressure  to  pass  through. 

This  will  not  be  done  here,  however. 

EXAMPLE.  —  As  an  example,  St.  Peter's  dome  will  be  given, 
with  the  dimensions  as  given  by  Scheffler  reduced  to  English 
measures.  The  dome  consists  in  its  upper  part,  as  will  be 
evident  from  the  figure,  of  two  domes  ;  the  lantern  resting  on 


DOMES. 


851 


the  two  is  assumed  to  have  one-third  of  its  weight  resting  on 
the  upper,  and  two-thirds  on  the 
lower  dome. 

Diameter  of  dome  =  diameter 
at  the  base  —  144  feet. 

Up  to  a  point  28.48  feet  above 
the  point  C  it  is  formed  of  a  single 
dome  11.84  feet  thick.  In  its 
upper  part,  on  the  other  hand, 
it  is  composed  of  two  domes 
whose  normal  distance  apart  is 
5.15  feet;  the  exterior  having  a 
thickness  of  2.56  feet,  and  the 
inner  of  4.13  feet  at  the  top  and 
5.15  feet  at  the  springing.  At 
the  top  of  these  two  domes  is  an 
opening  12.24  feet  radius,  sur- 
mounted by  a  cylindrical  lantern. 
The  magnitude  of  the  load  of  the 
lantern  on  the  dome  is  repre- 
sented on  the  figure  by  1.82  feet 
width  and  56.66  feet  height. 

Height  of  the  entablature 
ABCD  —  23.69. 

Width  of  ABCD  normal  to 
plane  of  paper  =  1.02  feet. 

Thickness  of  ABCD  —  10.30 
feet. 

Divide  the  exterior  dome  into 
nine  parts,  the  interior  into  eight 
of  a  uniform  circumferential  width  of   10.08   feet,  except  the 
first,  which  has  a  width  of  only  1.82  feet. 

Determine  whether  this  thickness  of  ABCD  is  sufficient  to 
keep  the  line  of  resistance  within  joint  AB. 


FIG.  295. 


852  APPLIED   MECHANICS. 


CHAPTER   X. 
THEORY  OF  ELASTICITY,  AND  APPLICATIONS. 

§  275.  Strains.  —  When  a  body  is  subjected  to  the  action 
of  external  forces,  and  in  consequence  of  this  undergoes  a 
change  of  form,  it  will  be  found  that  lines  drawn  within  the 
body  are  changed,  by  the  action  of  these  external  forces,  in 
length,  in  direction,  or  in  both  ;  and  the  entire  change  of  form 
of  the  body  may  be  correctly  described  by  describing  a  suffi- 
cient number  of  these  changes. 

If  we  join  two  points,  A  and  B,  of  a  body  before  the  exter- 
nal forces  are  applied,  and  find,  that,  after  the  application  of 
the  external  forces,  the  line  joining  the  same  two  points  of  the 
body  has  undergone  a  change  of  length  &(AB),  then  is  the  limit 

of  the  ratio  ,  as  AB  approaches  zero,  called  the  strain  of 


the  body  at  the  point  A  in  the  direction  AB. 

If  AB  +  &(AB)  >  AB,  the  strain  is  one  of  tension  ;  whereas, 
if  AB-\-  k(AB)  <  AB,  the  strain  is  one  of  compression. 

In  order  to  study  the  changes  of  form  of  the  body,  let  us 
assume  a  point  O  within  the  body  when  there  are  no  external 
forces  acting,  and  let  us  draw  through  this  point  three  rectangu- 
lar axes,  OX,  OY,  and  OZ,  and  assume  a  small  rectangular 
parallelopipedical  particle  whose  three  edges  are  OAt  OB,  and 


STRAINS. 


853 


/ 

• 

€t 

C   / 

d 

/^ 

_7/! 

~"~--J/ 

E 
/ 

/ 

c         StL. 

OC, 

f 

/ 

z 

u  & 

A 

J 
X 

^/ 

/ 

Y 

/ 

3 

OC,  and   let   us  examine  the  form   of   this   particle  after  the 

loads  are  applied ;    it  will  be 

found  that  the  edges  OA,  OB, 

and  OC  will    be   of   different 

lengths  from  what  they  were 

before,    and    that    the   angles 

AOB,  AOC,  and  BOC  will  no 

longer   be   right    angles,    but 

will  differ  slightly  from  90°. 

Let    the  parallelepiped  oabc- 

g  def  represent  the  form  and 

dimensions    of    the    particles 

after  the  external  forces  are 

applied.     Then  we  shall  have, 

if  c*,  ey,  and  cz  represent  the  strains  in  the  directions  OX>  OYt 

and  OZ  respectively,  that 

.  oa  —  OA 

*x  =  limit  of as  OA  approaches  zero, 

OA 

ey  =  limit  of as  OB  approaches  zero, 

«2  =  limit  of — — —  as  O  C  approaches  zero. 


FlGt 


In  the  figure,  *x  and  ez  are  tensile  strains,  and  <.y  is  a  com- 
pressive  strain. 

But  these  strains  do  not  represent  completely  the  distortion 
of  the  particle ;  for  the  plane  CEGD  has  slid  by  the  plane 
OABF  through  the  distance  ocu  the  distance  apart  of  these 
planes  being  OC,  and  the  plane  halfway  between  the  two  has  slid 
just  half  as  far,  so  that  the  amount  of  shearing,  or  the  shearing- 
strain  of  planes  parallel  to  XOY  in  the  direction  OX,  may  be 

represented  by  —    =  — -  nearly,  or  the  distortion  divided  by  the 


854  APPLIED   MECHANICS. 

distance  apart  of  these  planes.     This,  moreover,  is  the  tangent 
of   the   angle  occ^  or  the  tangent  of   the  angle  by  which  aoc 
differs  from  a  right  angle. 
If,  now,  we  let 
yzx  —  shearing-strain  in  a  plane  perpendicular  to  OZ  in  the 

direction  OX, 
yzy  =  shearing-strain  in  a  plane  perpendicular  to  OZ  in  the 

direction  OY, 
yyx  =  shearing-strain  in  a  plane  perpendicular  to  O Fin  the 

direction  OX, 
yyz  =  shearing-strain  in* a  plane  perpendicular  to  OY'm  the 

direction  OZ, 
yxz  —  shearing-strain  in  a  plane  perpendicular  to  OX  in  the 

direction  OZ, 
yxy  —  shearing-strain  in  a  plane  perpendicular  to  OX  in  the 

direction  OY, 

and  let  boc  = </>,  aoc  = &,  aob  — Y,  then  we  shall 

2       i£^w  2  2 

have 

yzx  =  — -1  =  tan  \f,,  yyz  —  tan  <£, 

f*i 

yzy  =  tan  </>,  y^  =  tan  i/r, 

y^=tanx,  y^=tanx. 

We  thus  have 

yzy  =  yya  =  tan  <f>, 

yxz  —  yzx  —  tan  \f/, 

yxy  =  7^  =  tan  x, 

three  very  important  equations. 

We  thus  have  to  determine  six  strains,  in  order  to  define 
completely  the  state  of  strain  in  a  body  at  a  given  point ;  viz., 
if  we  assume  three  rectangular  axes,  we  must  know  e^,  *y,  cz, 
yzy  —  ?yzt  y-zx-  =  y*z>  yxy  =  yyx,  three  normal  and  three  tangen- 
tial strains. 


STRAINS  IN    TERMS   OF  DISTORTIONS. 


855 


§  276.  Strains  in  Terms  of  Distortions. — For  the  sake  of 
clearness,  we  will  consider  first  only  the  strains  that  are  parallel 
to  the  z  plane ;  hence,  will  use  only  two  co-ordinate  axes,  OX 
and  OY,  as  shown  in  Fig.  297.  In  this  case  let  us  assume  a 
small  rectangular  particle,  acbd,  the  co-ordinates  of  one  corner 
of  which  are  #,  y,  and  of  the  other,  x  +  dx,  y  +  dy ;  this  being 


FIG.  297. 

the  case  before  the  load  is  applied.  Let  the  effect  of  the  load 
be  to  move  the  point  a  to  e,  and  b  to  f,  transforming  the 
rectangle  acbd  into  ekfl,  and  thus  changing  x,  y  respectively 
into  x  +  £,  y  +  q,  and  changing  x  -\-  dx,  y  +  dy  respectively 
into  x  -\-  Z  +  dx  -f-  d%,  y  +  r?  +  dy  +  drj.  Then  are  dx,  dy 
the  sides  of  the  particle  before  the  load  is  applied.  Then 
from  what  has  preceded  we  shall  have 


dx9 


£          —     [      • 

*>-      #' 


=  r,x  = 


dy    '    dx 
The  first  two  are  evident  at  once.     To  prove  the  third,  ob- 


856 


APPLIED    MECHANICS. 


serve  that  the  shearing-strain,  yxy  is  the  tangent  of  the  angle 
by  which  the  angle  kel  differs  from  a  right  angle ;  hence  it  is 
the  tangent  of  the  sum  of  the  angles  kem  and  len.  Now,  since 
these  angles  are  small,  we  may  take  the  sum  of  the  tangents 
as  nearly  equal  to  the  tangent  of  the  sum.  But 


km      dn 

tan  kem  =  —  =  —  nearly,  and 
em       doc          3 


.          In       dZ 
tan  /£#  =  —  =  —  nearly. 
en       dy 


Hence 


FIG.  298. 


In  the  general  case,  Fig.  298,  a  rectangular  parallelopipedi- 


STRAINS  IN   TERMS  OF  DISTORTIONS.  857 

cal  particle,  the  co-ordinates  of  one  corner  of  which  are  x,  y,  z, 
and  of  the  other,  x  -\-  dx,  y  +  dy,  z  -f-  dz  ;  this  being  the  case 
before  the  load  is  applied. 

Let  the  effect  of  the  load  be  to  change  x,  y,  z,  respec- 
tively, into  x  +  £,  y  +  ?7,  2  -f-  C,  and  to  change  x  +  dx,y  -\-  dy, 
z  -\-  dz,  into  (x  -\-  £)  -f-  (d*  +  d£),  (y  -f-  rj)  -|-  (*/j/  -f-  drf),  (2  +  C) 
+  (dz  +  */£).  Then  are  dx,  dy,  dz,  the  edges  of  the  particle 
before  the  load  is  applied. 

Then,  from  what  has  preceded,  we  shall  have 

dx  dy  dz ' 

dy       dx  dz       dx      yz  dz       dy 

The  first  three  will  be  evident  at  once.  As  to  the  last  three, 
the  proof  is  similar  to  that  just  used  in  the  case  of  two  co- 
ordinate axes. 


dy       dx  dz       dx  dz       dy 

§  277.  Determination  of  the  Strain  in  any  Given  Direc- 
tion. —  Suppose  we  are  required,  knowing  the  strains  e^,  ey,  cs, 
y*y,  yxz,  yyz,  to  determine  the  strain  in  a  direction  making  angles 
a,  ft  y,  with  OX,  O  Y,  OZ  respectively.  Assume  our  rectangu- 
lar parallelopipedical  particle  in  such  a  way  that  the  diagonal 
from  (x,  j,  z)  to  (x  +  dx,  y  +  dy,  z  +  dz)  shall  be  in  the 
required  direction,  and  call  the  length  of  this  diagonal  ds  (Fig. 
298)  ;  then  we  shall  have 


dx  ,  ^ 

COSa  =  —  -,  (2) 

ds 


858  APPLIED   MECHANICS. 

dz  .   . 

cosy  =  — .  (4) 

as 

Let  e  be  the  strain  in  the  required  direction ;  then  length  of 
diagonal  after  load  is  applied  will  be 

and  we  shall  have 


or 


(5) 


Now,  subtracting  (i)  from  (5),  and  neglecting  ^(ds) 
(drj)2y  and  (d£)2  as  being  very  small  compared  with  the  rest,  we 
have 


dx  ..        dy  ,         dz 
d&  =  -ft  +  ^  +  - 
^  «j-  dtf 

ids  —  dgcosa  H-  drjcosfi  +  d^cosy.          (7) 


(6) 
«j-  dtf 

or 


But 

^  =  ^  +  ^  +  <&,  (8) 

d&e  dy  dz 

A,  =      *  +  d-dy  +      *,  (9) 


dy  dz 


859 


Hence,  substituting  these,  we  have,  after  dividing  by  ds,  and 
observing  (2),  (3),  and  (4), 


or,  making  use  of  §  276,  we  have 

€  =   €^COS2a  +  tyCOS2  ft  +  €Z  COS2y  +  yyz  COS  (3  COS  y 

+  yxz  cos  a  cosy  +  y^,  cos  a  cos  A         (12) 

which  gives  us  the  strain  in  any  direction. 

It  can  be  shown  that  there  are  three  directions,  at  right 
angles  to  each  other,  that  give  the  maximum  strains  or  mini- 
mum strains  :  and  we  might  deduce  the  ellipsoid  of  strains,  in 
which  semi-diameters  of  the  ellipsoid  represent  the  strains  ;  but 
we  will  pass  on  to  the  consideration  of  the  stresses. 

§  278.  Stresses.  —  When  a  body  is  subjected  to  the  action 
of  external  forces,  if  we  imagine  a  plane  section  dividing  the 
body  into  two  parts,  the  force  with  which  one  part  of  the  body 
acts  upon  the  other  at  this  plane  is  called  the  stress  on  the 
plane  ;  and,  in  order  to  know  it  completely,  we  must  know  its 
distribution  and  its  direction  at  each  point  of  the  plane.  If  we 
consider  a  small  area  in  this  plane,  including  the  point  O,  and 
represent  the  stress  on  this  area  by/,  whereas  the  area  itself  is 

represented  by  a,  then  will  the  limit  of  —  ,  as  a  approaches  zero, 

be  the  intensity  of  the  stress  on  the  plane  under  consideration 
at  the  point  O.  Observe  that  we  cannot  speak  of  the  stress  at 
a  certain  point  of  a  body  unless  we  refer  it  to  a  certain  plane 
of  action  :  thus,  if  a  body  be  in  a  state  of  strain,  we  do  not 
attempt  to  analyze  all  the  molecular  forces  with  which  any  one 


36o 


APPLIED   MECHANICS. 


particle  is  acted  on  by  its  neighbors  :  but,  when  we  assume  a 
certain  plane  of  section  through  the  point,  the  stress  on  this 
plane  at  the  point  becomes  recognizable  in  magnitude  and 
direction  ;  and  what  the  magnitude  and  direction  of  the  stress 
at  the  given  point  is,  depends  upon  the  direction  of  the  plane 
section  chosen,  the  magnitude  and  direction  differing  for  differ- 
ent plane  sections  through  the  point. 

§  279.  Simple  Stress.  —  A  simple  stress  is  merely  a  pull 
or  a  thrust.     Assume  a  prismatic  body,  with  sides  parallel  to 
OX,  subjected  to  a  pull  in  the  direction  of   its 
length  ;  the  magnitude  of  the  pull  being  P.     As- 
^^  sume  first  a  plane  section  AA   normal  to   the 

T\  N  direction  of  P,  and  let  area  of  AA  be  A.     Then, 

/    if  px  represent  the  intensity  of   stress   at   any 
point  of  this  plane, 


This,  which  is  the  intensity  of  the  stress  as  dis- 
tributed over  a  plane  normal  to  its  direction,  may 
be  called  its  normal  intensity. 

On  the  other  hand,  if  we  desire  to  ascertain 

the  intensity  of  the  stress  on  the  oblique  plane  BB,  making  an 

angle  6  with  AA,  we  shall  have 


Area  BB  = 


cos  6 


Hence,  if  pr  represent  the  intensity  of  the  stress  on  this  plane 
in  the  direction  OX,  we  shall  have 


P 


(— ) 

\cos  0/ 


P 

=    —COS  0  = 


(I) 


If  we  resolve  this  into  two  components,  acting  respectively  nor- 


COMPOUND  STRESS.  86 1 


mal  and  tangential  to  BB,  and  if  we  denote  the  normal  intensity 
by  /„,  and  the  tangential  by  ph  we  shall  have 

/„  =  pr  cos  0  =  /*  cos2  0,  (2) 

pt  =  pr  sin  6  =  px  cos  6  sin  0.  (3) 

If,  now,  we  assume  another  oblique  plane  section,  perpen- 
dicular to  the  first,  we  shall  obtain  the  normal  //and  the  tan- 

gential //  stress  on  this  plane  by  substituting  for  0,  -  —  0  ; 

hence  we  obtain 

/„'=/*  sin  *0,  (4) 


(5) 
Hence  follows 

A'=A; 

or,  the  tangential  components  of  a  simple  stress  on  a  pair  of 
planes  at  right  angles  to  each  other  are  equal. 

§280.  Compound  Stress.  —  A  compound  stress  may  be 
accounted  to  be  the  resultant  of  a  set  of  simple  stresses,  and 
may  be  analyzed  into  different  groups  of  simple  stresses. 

PROPOSITION.  —  Whatever  be  the  external  forces  applied  to  a 
body,  if  through  any  point  we  pass  three  planes  of  section  at  right 
angles  to  each  other,  the  tangential  components  of  the  stress  on 
any  two  of  these  planes  in  directions  parallel  to  the  third  must 
be  of  equal  intensity. 

To  prove  this  proposition,  assume 
three  rectangular  axes,  origin  at  O,  and 
assume  a  rectangular  parallelopipedical 
particle,  as  shown  in  the  figure,  so 
small  that  we  may  without  appreciable 
error  assume  the  stress  on  any  one  of 
the  faces  to  be  the  same  as  that  on  the 

r  ,  ,  FIG.  300. 

opposite  face  ;   resolve  these  stresses, 

i.e.,  the  forces  exerted  upon  the  faces  of   the  particle  by  the 

other  parts  of  the  body,  into  components  parallel  to  the  axes. 


862  APPLIED   MECHANICS. 

Let  o>  =  intensity  of  normal  stress  on  the  x  plane, 
a-y  =  intensity  of  normal  stress  on  thej^  plane, 
o-2  =  intensity  of  normal  stress  on  the  z  plane, 
Txy  —  intensity  of  shearing-stress  on  x  plane  in  direction 

OY, 
rxz  =  intensity  of  shearing-stress  on  x  plane  in  direction 

OZ, 
-ryx  =  intensity  of  shearing-stress  on  y  plane  in  direction 

ox, 

ryz   =  intensity  of  shearing-stress  on  y  plane  in  direction 

OZ,J 
TZX  =  intensity  of  shearing-stress  on  z  plane  in  direction 

OX, 
rzy  =  intensity  of  shearing-stress  on  z  plane  in  direction 

OY.  J 

We  have  thus  apparently  nine  stresses,  which  must  be  given, 
in  order  to  define  the  stress  at  the  point  O  completely  ;  but  we 
will  now  proceed  to  prove  that 


In  the  figure,  the  only  ones  of  these  stresses  that  are  repre- 
sented are  the  following  :  — 

Xa.    ==  X         =   <T 


Zy    =  Z1yl    =  crz. 

The  other  four  are  omitted,  in  order  not  to  complicate  the 
figure. 

Now,  it  is  evident  that  the  total  normal  force  on  the  face 
AFGD  and  the  normal  force  on  the  face  OB  EC  balance  each 
other  independently,  and  likewise  with  the  other  normal  forces. 


GENERAL   REMARKS.  863 


The  only  forces  tending  to  cause  rotation  around  OZ  are 
the  equal  and  opposite  parallel  forces  rxy  (area  AFGD),  one  act- 
ing on  the  face  AFGD>  and  the  other  on  the  face  OB  EC  ;  and 
the  equal  and  opposite  forces  ryx  (area  FBEG),  one  acting  on 
the  face  FBEG,  and  the  other  on  the  face  CO  AD. 

The  first  pair  forms  a  couple  whose  moment  is  rxy  (area 
AFGD]  (xx),  and  the  second  has  the  moment  ryx  (area  FBEG) 


But 

Area  AFGD  =  (FA)  (zz,),     ar£a  FBEG  =  (FB)  (zzt) 

.'.      rxy(FA)  (**,)  (xxs)  =  r 

Cancelling  £#„  we  have 


But 

7^4  =  yyl     and 


^  (yy,) 


Q.  E.  D. 

In  a  similar  manner  we  can  prove 


ryz   =  rzy. 
GENERAL    REMARKS. 

From  what  precedes,  it  follows,  that,  when  we  have  the  six 
stresses 

(TX,  O-y,  <TZ,  Tj;y,  TXZ,  TyZ, 

or,  in  other  words,  the  normal  and  tangential  components  of 
the  stresses  on  three  planes  at  right  angles  to  each  other,  given, 
the  state  of  stress  at  that  point  is  entirely  determined  ;  and, 
when  these  are  given,  it  is  possible  to  determine  the  direction 
and  intensity  of  the  stress  on  any  given  plane. 


864  APPLIED   MECHANICS. 

Moreover,  if  three  rectangular  axes,  OX,  OY,  and  OZ,  be 
assumed,  and  the  direct  strains  along  these  axes  be  given,  and 
also  the  shearing-strain  about  these  axes,  then  the  direct  strain 
in  any  given  direction  can  be  determined,  and  also  the  shearing- 
strain  around  this  direction  as  an  axis. 

The  two  above-stated  propositions  furnish  two  of  the  funda- 
mental propositions  of  the  theory  of  elasticity,  the  third  being 
the  determination  of  the  relation  between  the  stresses  and  the 
strains. 

§  281.  Relations  Governing  the  Variation  of  the  Stresses 
at  Different  Points  of  a  Body.  —  If  we  assume  a  point  whose 
co-ordinates  are  (x,  y,  2),  and  a  small  parallelopipedical  particle 
having  this  point  and  the  point  (x  +  dx,  y  -f-  dy,  z  -f-  dz)  for  the 
extremities  of  its  diagonal,  we  shall  have,  for  the  edges  of  this 
particle,  dx,  dy,  dz,  respectively. 

Now  let  the  stresses  at  (x,  y,  z)  be 

vx,     a-y,     <TZ,     rxy,     rxz,     ryz  ; 

i.e,  <rx  denotes  the  normal  stress  on  any  plane  perpendicular  to 
OX,  and  passing  through  the  point  (x,  y,  z),  etc.  Then,  for 
the  planes  passing  through  (x  +  dx,  v  -\-  dy,  z  -\-  dz},  we  shall 
have  the  stresses 


d<Ty,    vz  +  d*z,    rxy  -f  drxy,    rxz  +  drxz,    ryz  -f-  dr 


yz. 


We  may  also  have  outside  forces  acting  upon  the  particle  in 
question  :  if  such  is  the  case,  let  the  components  of  the  result- 
ant external  force  along  the  axes  be  respectively 

Xdxdydz,          Ydxdydz,         Zdxdydz. 

•• 

Now  impose  the  conditions  of  equilibrium  between  all  the 
forces  acting  on  the  particle.  To  do  this,  place  equal  to  zero 
the  algebraic  sum  of  all  the  forces  parallel  to  each  of  the  axes 


RELATIONS  BETWEEN  STRESSES  AND   STRAINS.        865 

respectively,  the  moment  equations  having  already  been  incor- 
porated in  our  demonstration  that 


yz 


Hence  we  have  three  conditions  of  equilibrium,  as  follows  :  — 


+  (rxy+dTxy  —  rxy)dxdz+  (rxz+drxz  —  rxz)dxdy+Xdxdydz  =  o, 
(oy  +doy  —  oy)dxdz  +  (TXy+dTXy  —  Txy}dydz  +(TyZ  +dryz  —  Tyz  )dydx+  Ydxdydz  =  o, 
(oz  +daz  —  oz)dxdy  +  (ryz  +dryz  —ryz  }dxdz+  (rXz+dTxz  —  rxz)dzdy  +  Zdxdydz  —  o. 

Hence,  reducing,  and  dividing  by  dxdydz,  we  have 

^+^  +  ^+*=o,  (,) 

dx  dy          dz 

^,+  ^  +  ^  +  y= 

dx          dy          dz 


.+Z-a.  (3) 

dx          dy          dz 

If  the  particle  is  in  the  interior  of  the  body,  and  we  dis- 
regard its  weight,  then  X  =  Y  =  Z  =  o. 

Equations  (r),  (2),  and  (3)  give  the  necessary  relations  which 
the  variations  of  stress  from  point  to  point  must  satisfy  in  order 
that  the  conditions  of  equilibrium  may  be  fulfilled 

§  282.  Relations  between  the  Stresses  and  Strains.  — 
Before  proceeding  to  the  general  problems  of  composition  of 
stresses,  i.e.,  of  determining  from  a  sufficient  number  of  data 
the  stress  upon  any  plane,  we  will  first  discuss  the  relations 
between  the  stresses  and  the  strains  ;  and  we  will  confine  our- 
selves to  those  bodies  that  are  homogeneous,  and  of  the  same 
elasticity  throughout. 

From  what  we  have  already  seen,  if  to  a  straight  rod  whose 
cross-section  is  A  there  be  applied  a  pull  P  in  the  direction  of 


866  APPLIED  MECHANICS. 

its  length,  the  intensity  of  the  stress  on  the  cross-section  wi 

be 

P 


and,  if  E  be  the  tensile  modulus  of  elasticity  of  the  material  of 
the  rod,  the  strain  in  a  direction  at  right  angles  to  the  cross- 
section,  or,  in  other  words,  in  the  direction  of  the  pull,  will  be 


Now,  another  fact,  which  we  have  thus  far  taken  no  account 
of,  is,  that  although  there  is  no  stress  in  a  direction  at  right 
angles  to  the  pull,  or,  in  other  words,  although  a  section  at 
right  angles  to  the  above-stated  cross-section  will  have  no  stress 
upon  it,  yet  there  will  be  a  strain  in  all  directions  at  right  angles 
to  the  direction  of  the  pull  :  and  this  strain  will  be,  for  any  direc- 
tion at  right  angles  to  the  pull, 


being  of  the  opposite  kind  from  «  ;  thus,  if  e  is  extension,  €x  is 
compression,  and  vice  versa. 

Hence,  if,  at  any  point  O  of  such  a  rod,  we  assume  three 
rectangular  axes,  of  which  OX  is  in  the  direction  of  the  pull, 
and  we  use  the  notation  already  adopted,  we  shall  have 

P 

v*  =  -       <ry  =  <TZ  =  rxy  =  rxz  =  ryz  =  O, 


RELATIONS  BETWEEN  STRESSES  AND  STRAINS.        86/ 
MODULUS    OF    SHEARING    ELASTICITY. 

In  the  case  of  direct  tension  or  compression,  when  only  a 
simple  stress  is  applied,  we  have  denned  the  modulus  of  elas- 
ticity as  the  ratio  of  the  stress  to  the  strain  in  its  own  direction. 

Adopting  a  similar  definition  in  the  case  of  shearing,  we 
shall  have 


where  G  is  the  modulus  of  shearing  elasticity. 


GENERAL  RELATIONS  BETWEEN  STRESSES  AND  STRAINS. 

Whenever  a  compound  stress  acts  on  a  body  at  a  given 
point,  let  the  stresses  be 


then  we  shall  have,  for  the  strains. 


€*-£~mE~~m£'  ''  ~G* 

<Ty  I    ffx  I    <TZ  Txt 


-         _  _  - 

~  E        mE       mE'  "  G' 

This  enables  us  to  determine  the  strains  in  terms  of  the 
stresses,  as  soon  as  the  values  of  E,  G,  and  m  are  known  from 
experiment,  for  the  material  under  consideration. 

If,  on  the  other  hand,  the  stresses  be  required  in  terms  of 
the  strains,  we  can  consider  €*>  €y>  €z>  yxy>  yxz>  7?z>  as  known,  and 
determine  vxy  o-^  <rz,  rxyy  ryz,  rxa)  from  the  above  equations. 


868  APPLIED   MECHANICS. 


We  thus  obtain 


m 


=  <Ty >  (2) 

y  m  ^  ' 


(3) 


and,  by  solving  these  equations  for  the  stresses,  we  have 

T,  =  -?-£(<-  +  ^  +  f"  +  "\  (4) 

m  +  i    \  m  —  2      I 

<ry  =  -^—E(^  +  **  +  Cj>  +  C'Y  (5) 

/w  +  i\^  m  —  2      I  °' 


o1* 


w 
and  also 


rxy  =   67^,      (7)          rxz  =   Gyxz,      (8)          ryz  =   67^.      (9) 

These  equations  express  the  stresses  in  terms  of  the  strains. 
The  three  last  might  be  written  as  follows  (see  §  276) :  — 


<-> 


as  these  forms  are  often  convenient. 

§  283.  Case  when  o-z  =  o Inasmuch  as  there  are  many 

cases  in  practice  where  the  stress  is  all  parallel  to  one  plane, 
and  where,  consequently,  the  stress  on  any  plane  parallel  to 
this  plane  has  no  normal  component,  it  will  be  convenient  to 
have  the  reduced  forms  of  equations  (4),  (5),  and  (6)  which 
apply  in  this  case. 


VALUES  OF  E,   G,   AND  m  869 

Let  the  plane  to  which  the  stresses  are  parallel  be  the  Z 
plane  ;  then  <rg  —  o.     Then  equation  (6)  becomes 


m-2 


and,  substituting  this  value  of  ez  in  (4)  and  (5),  and  reducing,  we 
obtain 


m 


m  + 

-^—£(fy  +  *4^?\  (2) 

m  +  \     \y       m  —  i/  v/ 


ivhich  are  the  required  forms. 

The  other  three  equations,  viz.,  — 


Gy 


yg, 


remain  the  same  as  before. 

§284.  Values  of  E,  G,  and  m.  —  These  three  constants 
need  to  be  known,  to  use  the  relations  developed  above. 

i°.  As  to  E,  this  is  the  modulus  of  elasticity  for  tension, 
and  has  been  determined  experimentally  for  the  various  mate- 
rials, as  has  been  already  explained.  Moreover,  it  has  also  been 
shown  experimentally,  that,  with  moderate  loads,  the  modulus 
of  elasticity  for  compression  is,  nearly  identical  with  that  for 
tension  in  cast-iron,  wrought-iron,  and  steel. 

2°.  As  to  m,  in  those  few  applications  that  Professor  Ran- 
kine  gives  of  his  theory  of  internal  stress,  such  as  the  case  of 
combined  twisting  and  bending,  he  determines  the  greatest  in- 
tensity of  the  stress  acting  ;  and  his  criterion  is,  that  this  shall 
be  kept  within  the  working-strength  of  the  material.  This  is 
-equivalent  to  assuming  m  =  oo.  The  more  modern  writers, 


8/0  APPLIED  MECHANICS. 

such  as  Grashof  and  others,  take  account  of  the  fact  that  m  has 
a  finite  value,  and  make  their  criterion  that  the  greatest  strain 
shall  be  kept  within  the  quotient  obtained  by  dividing  the  work- 
ing-strength by  the  modulus  of  elasticity  of  the  material. 

Thus,  if  f  is  the  working-strength,  and  o-T  the  greatest 
stress,  and  cx  the  greatest  strain,  Rankine's  criterion  of  safety 
is 


whereas  the  more  modern  criterion  is 


The  resulting  formulae  differ  in  each  case;  and,  as  has  been 
stated,  those  of  Rankine  could  be  derived  from  the  more  gen- 
eral ones  by  making 

i 

-  —  o    or    m  =  oo, 

which  is  never  the  case. 

As  to  the  value  of  m,  but  few  experiments  have  been  made. 
Those  of  Wertheim  give,  for  brass,  2.94  ;  for  wrought-iron,  3.64. 

The  values  m  =  3  and  m  =  4  are  those  most  commonly 
adopted,  so  that 

—  =  -  .1  =  I 

m       3  m       4* 

3°.  The  value  of  G,  the  shearing-modulus  of  elasticity,  i.e.,, 
the  ratio  of  the  stress  to  the  strain  for  shearing,  has  been 
determined  experimentally,  and  has  generally  been  found  to  be 
about  two-fifths  that  for  tension. 

According  to  the  theory  of  elasticity,  we  must  have 


as  may  be  proved  as  follows  :  — 


VALUES  OF  E,   G,   AND   m.  8;  I 

Assume  a  square  particle  whose  side  is  #,  and  let  a  simple 
normal  stress  a-  be  applied  at  the  face  AB ;  then  we 
shall  have,  on  the  planes  BD  and  AC,  a  shearing-stress  |d 

(§  279) 

T  =  o-  sin  45°  cos  45°  =  ^<r. 
On  the  other  hand,  if  we  let 


r 

I 

FIG.  301. 

the  strain  of  the  particle  in  the  direction  AD  will  be  e,  while 

that  in  the  direction  AB  will  be  —  —  ;  hence  the  particle  will 

m 

become  a  rectangle,  the  side  AD  changing  its  length  from  a  to 
a  -f-  ac,  and  side  AB  changing  from  a  to  a . 

The  diagonals  will  no  longer  be  at  right  angles  to  each 
other ;  and,  if  we  denote  by  a  the  angle  by  which  their  angle 
differs  from  a  right  angle,  we  shall  have,  for  the  shearing-strain 
on  the  planes  A  C  and  BD, 

y  =  tana. 

But,  after  the  distortion,  the  angle  ADB  will  become 


i   —  tan-        a i  —  — 

.  _  2 w  m . 


therefore,  dividing,  and  carrying  the  division  only  to  terms  of 
the  first  degree,  we  have 

1  —  2  tan-  =  i  —  [  i  -f  —  \ 
2  \         m/ 


A     stanlr)  -^-c. 


8/2  APPLIED   MECHANICS. 


But 


y  =  tan  a  =  2  tan  -  nearly 


T  _       if0"       _  i      m     /o-\ . 
y       m  "i"  Ic       2  #2  •+•  J  V €  / 


M 

but 

I  =  £      and     -  = 

7  € 

I  m  r, 


§  285.  Conjugate  Stresses.  —  If  the  stress  on  a  given  plane 
at  a  given  point  of  a  body  be  in  a  given  direction,  the  stress  at 
the  same  point  on  a  plane  parallel  to  that  direction  will  be 
parallel  to  the  given  plane.  Let  YO  Y  represent,  in  section,  a 
given  plane,  and  let  the  stress  on  that  plane  be  in  the  direction 
XOX. 

Consider  a  small  prism  ABCD  within  a  body,  the  sides  of 
whose  base  are  parallel  respectively  to  XOX  and  YO  Y.  The 
forces  on  the  plane  AB  are  counterbalanced  by  the  forces 
on  the  plane  DC ;  the  resultants  of  each  of  these  sets  being 
equal  and  opposite,  and  acting  along  a  line  passing  through  O. 
Hence  the  forces  acting  on  the  planes  AD  and  BC  must  be  bal- 
anced entirely  independently  of  any  of  the  forces  on  AB  or 
DC:  and  this  can  be  the  case  only  when  their  direction  is  paral- 
lel to  YOY ';  for  otherwise  their  resultants,  though  equal  in 
magnitude  and  opposite  in  direction,  would  not  be  directly 
opposite,  but  would  form  a  couple,  and,  as  there  is  no  equal  and 
opposite  couple  furnished  by  the  forces  on  the  other  faces,  equi- 
librium could  not  exist  under  this  supposition. 

§286.  Composition  of  Stresses.  —  The  general  problem  of 
the  composition  of  stresses  may  be  stated  as  follows :  — 


PROBLEM. 


8/3 


Knowing  the  stresses  at  a  given  point  of  a  strained  body 
on  three  planes  passing  through  that  point,  to  find  the  stress  at 
the  same  point  on  any  other  plane,  also  passing  through  the 
same  point.  The  stresses  on  the  three  given  planes  are  not 
entirely  independent ;  in  other  words,  we  could  not  give  the 
stresses  on  these  three  planes,  in  magnitude  and  direction,  at 
random,  and  expect  to  find  the  problem  a  possible  one.  Thus, 
suppose  that  the  planes  are  at  right  angles  to  each  other,  we 
have  already  seen  that  we  have  the  right  to  give  their  three 
normal  components,  a-*,  o>,  and  <rz,  and  the  three  tangential, 
TV?*  T?Z,  and  rXZ9  and  that  ryx  =  rxy,  etc.  We  will  now  proceed 
to  special  cases. 

§  287.  Problem.  —  Given  the  three  planes  of  action  of  the 
stress  as  the  x,  y,  and  z  plane  respectively,  and  given  the  nor- 
mal and  tangential  components  of  the  stresses  on  these  planes, 
viz.,  a-^  <ry,  crz,  rxy,  TXZ,  and  ryz,  to  find  the  intensity  and  direction 
of  the  stress  on  a  plane  whose  normal  makes  with  OX,  OY, 
and  OZ  the  angles  a,  ft  and  7  respectively,  where,  of  course, 

COS2  a  +  COS2  fi  4"  COS 2  7  =   I. 

Draw  the  line  ON9  making  angles  a,  ft  and  7  with  OX,  O  Y, 
;and  OZ  respectively  ;  then  draw 
near  0  the  plane  ABC  perpen- 
dicular to  ON.  It  has  the  direc- 
tion of  the  required  plane,  and 
cuts  off  intercepts  OA,  OB,  and 
OC  on  the  axes ;  and,  moreover, 
we  shall  have,  from  trigonometry, 
the  relations, 

Area  BOC  =  (ABC)  cos  a, 

Area  AOC  =  (ABC)  cos  ft 

Area.  A  OB  =  (ABC)  cosy. 


FIG.  302. 


Now  consider  the  conditions  of  equilibrium  of  the  tetrahe- 
dron OABC.     The  stress  on  ABC  must  be  equal  and  directly 


8/4  APPLIED  MECHANICS. 

opposed  to  the  resultant  of  the  stresses  on  the  three  faces 
AOCy  BOC,  and  AOB.  Now  let  us  proceed  to  find  this  result- 
ant. 

In  the  direction  OX  we  have  the  force 

<rx(BOC)  +  rxz(AOB)  +  rxy(AOC) 

=  (A£C)(o-xcosa  +  r^cos/3  4-  r^cosy). 


Lay  off  6>Z>  to  represent  this  quantity.    In  the  same  way  repre- 
sent the  force  in  the  direction  O  Y  by 

OE  =  vy(AOC)  4-  ryz(BOA)  +  rxy(BOC) 

Cosp  4-  r^cosa  4-  r^cosy)^ 


and  that  in  the  direction  OZ  by 


Now  compound  these  three  forces,  and  we  have,  as  resultant 
force, 

y?  =  OG  = 

and  as  resultant  intensity 


0-=  —  +  O£2  -4- 


r^cos/3  -f 
+  (a-yCOS/3  + 

4-  (o-^cosy  +  r^cosa  + 
=  \f\o-j?  COS2  a  +  <r/COS2/?  +  trz2COS2y 

+  T*./(cos2a  +  cos2/?)  4-  rjr/(cos2a  +  cos2y) 
+  Tya2(cos2/3  4-  cos2y)  4-  2o--r(T^cosj8  4-  r^cos 

-h  2<Ty  (T^,  COS  a  4-  Tyz  COS  y)  COS 
4-  20-3(1-.^  COS  a  +  ryz  COS  y8)  COS  y  +  2rjeyrxz  COS  y8  COS  y 

4-  2r^Ty2  COS  a  COS  y  4-   2x^x^2  COS  a  COS  ft\ 


STRESSES  PARALLEL    TO  A   PLANE. 


8/5 


the  direction  being  given  by  the  angles,  ar,  f!r,  and  yr,  where 


§  288.  Stresses  Parallel  to  a  Plane.  —  To  solve  the  same 
problem  when  there  is  no  stress  in  the  direction  OZt  and  when 
the  new  plane  is  perpendicular  to  XO  Yy  or,  in  other  words,  in 
the  case  when  the  planes  of  action  are  all  perpendicular  to  one 
plane,  to  which  the  stresses  are  all  parallel  :  we  then  have 


<jz  =  rxz  =  Tyz  =  O     and     fi  =  90°  —  a, 


and  hence 


cr  =  \a-x2CQS2a  -f-  oj,2sin2a  -f-  rxy2 


-f-  ay)  rxy  COS  a  sin  a. 


FIG  0 


Or  we  may  proceed  as  follows  :  — 

Let  the  normal  intensity  of  the  stress  on  the  x  plane  (i.e.,. 
that  perpendicular  to  OX}  be  cr*,  that  on  the 
y  plane  o>,  and  the  tangential  intensity  rxy. 
Let  ON  be  the  direction  of  the  normal  to 
the  plane  on  which  the  stress  is  to  be  deter- 
mined, and  let  the  angle  XON  —  a.  Then 
let  the  plane  AB  be  drawn  perpendicular  to 
ON,  and  let  us  consider  the  equilibrium  of 
the  forces  exerted  by  the  other  parts  of  the 
body  upon  the  triangular  prism  whose  base  is  ABO  and  alti- 
tude unity. 

If  we  compound  the  forces  acting  on  the  faces  AO  and  OB, 
we  shall  have,  in  their  resultant,  the  total  force  on  the  face  AB- 
in  magnitude  and  direction.  Moreover,  we  have  the  relations,, 

Area  OB  =  area  AB  cos  a    and    Area  OA  =  area  AB  sin  a. 
Force  acting  on  OB  in  direction  OX  =  <rx(OJ3), 
Force  acting  on  OB  in  direction  OY  =  rxy(OB), 
Force  acting  on  OA  in  direction  OX  =  r^OA^ 
Force  acting  on  OA  in  direction  OY  =  <ry(OA)* 


876  APPLIED  MECHANICS. 

Hence,  if  we  lay  off 

OD  =  <rx(OB)  +  rxy(OA)     and     OC  =  <ry(OA)  -f-  r 

then  will  <9Z>  represent  the  total  force  acting  in  the  direction 
OXt  ^and  OC  will  represent  the  total  force  acting  in  the  direc- 
tion O  Y. 

Compounding  these,  we  shall  have  OE  as  the  resultant  total 

OF? 
force  on  the  face  AB,  and  —  —  will  represent  its  intensity. 


To  deduce  the  analytical  values,  we  have 

OD  =  <rx(OB)  -f-  rxy(OA)   =  (^Xo-^COSa  +  r 

OC  =  <ry(OA)   4-  rxy(OB)  =  (AB)(<rysma    +  r^COSa) 


OE  =  <JOD2  +  OC 


Tjy,sina)2  -f  (o-^sina  +  T^,cosa)2 
AB\\<Tj?  COS  2  a  4-  cr/  sin2  a  +  2r^y  COS  a  sin  a(a-^  +  <ry) 

-f  sin2 


Or,  if  oy  represent  the  resultant  intensity  on  the  plane  AB,  and 
OT  the  angle  this  resultant  makes  with  OX,  we  shall  have 

oy  =  V^<rjr2COS2a  -f  <r/sm2a 

+  2r^(o-^  -f  ory)  COS  a  sin  a  +  rxf  \,      (  I  ) 

and 

=  and 


Moreover,,  it  is  sometimes  desirable  to  resolve  the  stress  into 
normal  and  tangential  components.  If  this  be  done,  and  if  a- 
and  T  represent  respectively  the  normal  and  tangential  com- 
ponents, we  shall  have 


PRINCIPAL   STRESSES. 


but 

OF  =  O&cosa  +  £Dsina    and    EF  =  £jDcosa  —  ODsina 

OD  OC  . 

.'.       cr  =    -  COS  a   +    -  —Sin  a 

AB  AB 

=  a-xCOS2a  +  <TySin2a  •+•  2Tjcy  COS  a  sin  a      (2) 

and 


OD  . 

T  =  -  COS  a   —    -  -  Sin  a 


=  (a-y  —  crx)  COS  a  sin  a  +  T*y(cos2a  —  sin2a) 

(cTy    —    <TX\ 
—  -  j  Sin  2a  +  T^,COS  2a.    (3) 

§289.  Principal  Stresses.  —  It  will  next  be  shown,  that, 
whatever  be  the  state  of  stress  in  a  body,  provided  the  stresses 
are  all  parallel  to  one  plane,  the  planes  of  action  being  all  taken 
perpendicular  to  this  plane,  there  are  always  two  planes,  at  right 
angles  to  each  other,  on  which  there  is  no  tangential  stress  ; 
these  two  planes  being  called  the  planes  of  principal  stress,  the 
stress  on  one  of  these  planes  being  greater,  and  the  other  less, 
than  that  on  any  other  plane  through  the  same  point. 

To  prove  the  above,  it  will  be  necessary  only  in  the  last 
case,  which  is  a  perfectly  general  one,  to  determine  for  what 
values  of  a  the  value  of  r  is  zer.o,  and  whether  these  values  of  <*. 
are  always  possible.  We  have 


T  =    — sin  20.  +  rxy  cos  20. : 

and,  if  we  put  this  equal  to  zero,  we  have 
sin  20,  2TrV 


COS  2tt 


tan  20.  — 


and  this  gives  us,  for  all  values  of  o^,  cryy  and  rxy,  two  possible 
values  for  2a,  differing  from  each  other  by  1 80°,  hence  two 
values  for  a  differing  by  90°.  Hence  follows  the  first  part  of; 
the  proposition. 


$78  APPLIED   MECHANICS. 

The  latter  part  —  that  these  are  the  planes  of  the  greatest 
and  least  stresses  —  will  be  shown  by  differentiating  the  value 
of  (rr2y  and  putting  the  first  differential  co-efficient  equal  to  zero  ; 
and,  as  this  gives  us 

—  2<rx  cos  a  sin  a  -j-  2oy*  cos  a  sin  OL 
4-  *Txy(?x  +  o>)(cos2a  —  sin2  a) 

=  2  (<rx  -f-  (Ty)  \  (cry  —  a-x)  COS  a  sin  a  -f  T^y(cos2  a  —  sin2  a)  \ 


therefore  we  have  the  same  condition  for  the  maximum  and 
minimum  stresses  as  we  have  for  the  planes  of  no  tangential 
stress. 

It  follows  that  the  determination  of  the  greatest  and  least 
stresses  at  any  one  point  of  a  body  is  identical  with  the  deter- 
mination of  the  principal  stresses ;  and  it  will  be  necessary, 
whenever  the  stresses  on  any  two  planes  are  given,  to  be  able 
to  determine  the  principal  stresses,  as  one  of  these  is  the 
greatest  stress  at  that  point  of  the  body,  and  the  other  the 
least. 

§  290.  Determination  of  Principal  Stresses.  —  When  the 
stress  is  all  parallel  to  one  plane,  viz.,  the  z  plane,  and  when 
the  stresses  on  two  planes  at  right  angles  to  each  other  are 
given,  i.e.,  their  normal  and  tangential  components,  we  may  be 
re'quired  to  determine  the  principal  stresses.  Proceed  as  fol- 
lows :  Given  normal  stresses  on  X  and  Y  planes  respectively, 
<rx  and  cry,  and  tangential  stress  on  each  plane  rxyt  to  find  prin- 
cipal stresses. 

From  §  288  we  have,  for  a  plane  whose  normal  makes  an 
angle  a  with  OX, 

°"r  =  Vo^/COS2  a  +  <r/sin2a  -f  2Txy(<rx  -f-  oy)cOSasina  -f-  T*-/,     (l) 
o*  =  crxCOS2a  +  ay  sin2  a  -|-  2rxy  COS  a  sin  a,  (2) 

T  =  i>y  (cos2  a  —  sin2  a)  —  (<rx  —  oy) cos  a  sin  a,      (3) 


DETERMINATION  OF  PRINCIPAL   STRESSES.  879 

or 

T  =  rxy  cos  2a sin  2a.  (4) 

Now,  the  condition  that  the  plane  shall  be  a  plane  of  prin- 
cipal stress  is,  that  T  =  o.     Hence  write 

Tjrj,(cos2a  —  sin2  a)  —  (<rx  —  o-^)  cos  a  sin  a  =  o, 

find  a,  and  substitute  its  value  in  (2),  and  we  shall  have  the 
principal  stresses.  The  operation  may  be  performed  as  fol- 
lows ;  viz.,  — 

From  (3)  we  have 

<TX    <Ty 

(a)      2  COS2  a  —  I  =  COS  asm  a 


1  (          <r*  -  °>  .      ) 
.*.     COS2 a  =  -<  I  H -COS  a  Sin  a  > 

2  (  rxy  \ 


>xy 
(h\       1—2  Sin2  a   = COS  a  sin  a 

I    (  <TX    —     <Ty  \ 

.*.      Sin2  a  =  -  \  I COS  a  Sin  a  > . 

2(  rxy 

Hence 

<r*  +  o>        COS  a  Sin  a  j  <rx2  —  2arxa-y 


or 

<r*  +  °>        cos  «  sin  a 

2 


(  \ , 

(<r*-o>) 2 


But  we  have,  since  (4)  equals  zero, 
tan  2a  = — 


sin  2a  =  2  sin  a  cos  a 


3 So  APPLIED   MECHANICS. 

Hence  substitute  for  cos  a  sin  a  its  value,  and 

ov  -J-  (TV 


yY  +    4T,/,  (5) 

which  gives  us  the  magnitudes  of  the  principal  stresses ;  the 
plus  sign  corresponding  to  the  greater,  and  the  minus  sign  to 
the  less. 

EXAMPLES. 

i.  Let,  in  the  last  section,  <ry  =  o,  and  find  the  principal  stresses. 
Here  we  have 

2rxy 

tan  2<x  =  — - 
o> 

and 


=^±* 

2  2 


2.  Given  two  principal  stresses,  to  find  the  stress  on  a  plane  whose 
normal  makes  an  angle  a  with  OX. 

In  this  case  rxy  =  o. 

Hence  we  have  the  case  of  §  288,  with  the  reduction  of  making 
txy  =  o.  We  may  therefore  obtain  the  result  by  substitution  in  the 
results  of  §  288,  or  we  may  proceed  as  follows  :  — 

(a)  Find  stress  on  new  plane  in  direction  OX;  this  will  be,  §  279, 

<TJ.  cos  a. 
(b\  Find  stress  on  new  plane  in  direction  OY;  this  will  be,  §  279, 

cry  sin  a. 
(i)  Compound  the  two,  and  the  resultant  is 


(Tr  =  Vo-/COS2a  +  o-/sin2a.  (l) 

(d)  Normal  component  of  <rx  cos  a  is 

(rx  COS2  a. 

(i)  Normal  component  of  <ry  sin  a  is 

cr  sin2  a. 


ELLIPSE   OF  STRESS. 


88 1 


(/)  Add,  and  we  have,  for  normal  stress, 

<rn  =  crx  COS2  a  +  cry  sin2  a. 
(g)  Tangential  component  of  vx  cos  a  is 
— <rx  cos  a  sin  a. 

(h)  Tangential  component  of  cry  sin  a  is 
-\-vy  cos  a  sin  a. 

(k)  Add,  and  we  have,  for  tangential  stress, 
T  =   (a-y  —  <rx)  COS  a  sin  a. 


(3) 


§291.  Ellipse  of  Stress — In  the  case  above,  i.e.,  when 
the  two  principal  stresses  are  <rx  and  <ry  respectively,  if  we 
represent  them  graphically  by  OA  = 
<rx  and  OB  —  o-y)  and  let  CD  be  the 
plane  on  which  the  stress  is  required, 
its  normal  making  with  OX  the  angle 
XON  =  a,  then,  from  what  has  been 
shown,  if  OR  represent  the  intensity  of 
the  resultant  stress  on  this  plane,  we 
shall  have 


OR  =  <rr  =  VCT/  COS2  a  -f-  <r/  sin2  a  ; 

and,  moreover, 

&51  =  <jx  cos  a,        &F  =  <ry  sin  a. 

If  we  denote  these  by  x  and  y  respectively,  letting  (x,  y)  be 
the  point  R,  i.e.,  the  extremity  of  the  line  representing  the 
stress  on  AB,  then 


X   = 


-    O'- 


cos2  a 


and 


y  =  <Ty  sm  a, 


—     =  sn2  a 


f 


382  APPLIED   MECHANICS. 

which  is  the  equation  of  an  ellipse  whose  semi-axes  are  a-x  and 
<ry  respectively ;  hence  the  stress  on  any  plane  will  be  repre- 
sented by  some  semi-diameter  of  the  ellipse. 


SPECIAL  CASES. 

I.  When  the  two  given  stresses  are  equal,  or  <rx  =  o-y,  then 


o>  =  \(rx2COS2a  +  <Tysna  =  <rx, 

and 

vx  cos  a 
cos  OT  =  -  =  cos  a        and        smtv  =  sina/ 


therefore  the  stress  is  of  the  same  intensity  on  all  planes,  and 
always  normal  to  the  plane. 

II.  When  the  two  given  stresses  are  equal  in  magnitude 
but  opposite  in  sign,  or  <ry  =  —  <rx,  then 

oy  =  o> 

But 

cos  OT  =  cos  a        and        sinar  =  —sina, 
hence 

OT  =    —  a/ 

therefore  the  stress  on  any  plane  whose  normal  makes  an  angle 
a  with  OX  is  of  the  same  intensity  <rx,  but  makes  an  angle 
equal  to  a  with  OX  on  the  side  opposite  to  that  of  the  normal 
to  the  plane. 

PROBLEM.  —  A  pair  of  principal  stresses  being  given,  to 
find  the  positions  of  the  planes  on  which  the  shear  is  greatest. 

Solution.  —  Let  T  =  (<ry  —  o>)  sin  a  cos  a  =  max. 

Therefore  differentiate,  and 

cos2  a  —  sin2  a  =  o 

A     COS  a  =    ±sina  .'.     a  =  45°  or   135°. 


SPECIAL   MODES  OF  SOLUTION  OF  SOME  PROBLEMS.    883 

§  292.  Some  Special  Modes  of  Solution  of  some  Prob- 
lems. —  The  case  where  two  principal  stresses,  <rx  and  <ry,  are 
given,  to  find  the  stress  on  any  plane  whose  normal  makes  an 
angle  a  with  OX,  may  be  solved  as  follows,  graphically :  — 

Let,  Fig.  302,  o>  =  OA,  and  oy  =  OB.     Let  XON  =  a. 

Now, 

o~v  — —  * ~l~  . 

*  2  2          ' 

_  <?y  +  <*x         oy  —  Q>- 
ov  —  —•  . 


Hence,  instead  of  proceeding  at  once  to  find  the  resultant 
stress  on  CD  due  to  the  action  of  <rx  and  oy,  we  may  first  find 
that  due  to  the  action  of  the  two  equal  principal  stresses  of  the 
same  kind, 


then  that  due  to  the  pair 


j 
and 


2  2 

and  then  the  resultant  of  these  two  resultants. 

The  first  resultant  will  be  evidently  laid  off   on    ON,  and 

equal  in  magnitude  to  —  -  —  ;  hence  let  OM  =  —  -  -,  and 

2  2 

OM  will  be  the  first  resultant. 

The   second  resultant  will  be  of  magnitude  — 
will  have  a  direction  MR  such  that  the  angle  NMS  =  SMR. 

Hence,  laying  off  this  angle,  and  making  MR  = 

we  shall  have  for  the  final  resultant,  OR,  as  before. 

This  construction  will  be  useful  in  the  following  case:  — 
To  find  the  most  oblique  stress,  we  must  find  for  what 

value  of   a  the  angle  MOR  is   greatest.     This  will   be   made 


884  APPLIED   MECHANICS. 

evident  if  we  observe,  that,  for  all  positions  of  the  plane,  the 
triangle  OMR  has  always  OM  =  "y  2  "* ,  and  MR  =  "y  ~  ** -r 

both  of  constant  length.  Hence,  if,  with  M  as  a  centre  and 
MR  as  a  radius,  a  circle  were  described,  and  a  tangent  were 
drawn  from  O  to  this  circle,  the  point  of  tangency  being  taken 
for  R,  then  will  OR  be  the  most  oblique  stress ;  i.e.,  the 
stress  is  most  oblique  when  ORM  =  90°.  Therefore  greatest 
obliquity  = 


°>  — 
sin-1  -z-— - 


§  293.  Converse  of  the  Ellipse  of  Stress.  —  The  converse 
of  the  ellipse  of  stress  would  be  the  following  problem  :  Given 
any  two  planes  passing  through  the  point  in  question ;  given 
the  intensities  and  directions  of  the  stresses  on  these  planes,, 
—  to  find  the  principal  stresses  in  magnitude  and  in  direction. 

The  first  step  to  be  taken  is,  to  assure  ourselves  that  the 
conditions  are  not  incompatible,  as  they  are  liable  to  be  if  the 
planes  and  stresses  are  taken  at  random.  The  test  of  this 
question  is,  to  resolve  each  stress  into  two  components,  respec- 
tively parallel  to  the  two  planes ;  and,  if  the  conditions  are  not 
inconsistent,  the  component  of  each  stress  along  the  plane  on 
which  it  acts  must  be  equal.  The  proof  of  this  statement  can 
be  made  in  a  similar  way  to  that  used  in  proving  that  the 
intensities  of  the  shearing-stresses  on  two  planes  at  right  angles 
to  each  other  are  equal.  If,  upon  applying  this  test,  we  find 
that  the  conditions  are  not  inconsistent,  we  may  proceed  as 
follows  :  — 

Suppose  CD  (Fig.  304)  were  the  given  plane,  and  OR  the 
stress  upon  it,  and  suppose  the  position  of  the  principal  axes, 
OX  and  O  Y,  and,  indeed,  all  the  rest  of  the  figure,  were  absent, 
i.e.,  not  known.  Now,  we  can  easily  draw  the  normal  ON; 
and,  if  we  could  determine  upon  it  the  point  M  such  that  OM 


CONVERSE   OF   THE  ELLIPSE   OF  STRESS.  885 

should  be  one-half  the  sum  of  the  principal  stresses,  we  should 
be  able  to  reproduce  the  whole  figure.  Hence  we  will  devote 
ourselves  to  the  determination  of  the  position  of  the  point  M. 

Let  OR  =  p  =  stress  on  plane  CD. 

Let  stress  on  the  other  given  plane  be/x. 

Let  NOR  =  0  =  obliquity  of  /. 

Let  0,  —  obliquity  of  /,. 

Then  we  have 

MR*  =  OR2  +  OM2  -  2OM.  ORca&Ot 

or,  if  or*  and  <ry  denote  the  (unknown)  magnitudes  of  the  prin- 
cipal stresses, 


From  the  triangle  constructed  in  the  same  way,  with  the 
stress  on  the  other  plane,  we  should  have 

(Tx  +  o>\2 


Hence,  by  subtraction, 

/--/,»  =  2^  "^  ^  (/  cos  0  -  /x  cos  0r)  (3) 


= 

2  "   2(/COS0  —  /.COS0,)   * 

Having  thus  found  ^       ^  we  can  next  find,  from  either 

2 

(i)  or  (2),  the  value  of  —  -  -. 

Now,  therefore,  we  know  CM/  and  MR,  and  hence  we  can  lay 
voff  this  value  of  OM,  and  complete  the  triangle  OMR  ;  then 


886  APPLIED  MECHANICS. 

bisect  the  angle  NMR,  and  the  line  MS  is  parallel  to  the  axis 
of  greater  principal  stress.  Hence  draw  O  Y  parallel  to  MS, 
and  OX  perpendicular  to  OY,  and  lay  off  on  OY 


=  cry=  OM+  MR, 
and  on  OX 

OA  =  o-x=  OM—MR, 

and  the  problem  is  solved. 

§  294.  Rankine's  Graphical  Solution.  —  The  following  is 
Rankine's  graphical  solution  of  the  preceding  problem  : 

Draw  the  straight  line  ON,  Fig.  305,  and  then  lay  off  angle 
NOR  =  6,  and   angle   NOR, 
=  0l9  also  OR=p  and   OR,    0 
=  /,.     Then  join  R  and  RIf 
and  bisect  RRl  by  a  perpen- 
dicular SM.     From  the  point 
Mt  where  this  meets  ON,  draw  MR  and  MR^  .     Then  will 


/.  o-x  =  OM  -  MR,    (3)     and  <ry  =  OM  +  MR  ;       (4) 

and  the  angles  made  by  O  Y  with  6Wand  ONl  ,  respectively, 
will  be 

NMR  NMR, 

90    ~a  =  ~~2~~'        ^  9°    ~^1  =  --  2  —  * 

A  comparison   of  this  figure  with  the  triangle   OMR  of 
§  291  will  show  that  this  is  merely  a  graphical  construction  for: 


RANKINE'S   GRAPHICAL   SOLUTION. 


887 


the  analytical  solution   given  in  §  293  ;  and  the  equations  of 
that  article  can  readily  be  deduced  from  the  figure  given  above. 


SPECIAL   CASES. 

(a)  When  the  two  given  planes  are  at  right  angles  to  each 
other. — In  this  case  the  tangential  components  of  OR  and  ORl 
are  equal,  and  hence  (Fig.  306)  aR1  =  bR. 

Hence  the  figure  be- 
comes that  shown  where  o 
RR,  is  parallel  to  ON\  and 
if  we  let  Ob=pnyOa=pni, 
and  bR  =  aRl  =pt,  we  shall 
have 


whence  we  readily  obtain  <rx  and  o-tf,  and  then  a  and  or, ,  just  as 
before. 

(b)  When  the  two  given  stresses  are  conjugate  (see  §  285). — In 
this  case  the  obliquities  of  the  two  stresses  are  the  same,  and 
the  figure  becomes  Fig.  307. 

We  then  obtain  M 


COS  u 


FIG.  307. 


888  APPLIED  MECHANICS. 


=  MR  =  V(SRY  +  (MS)9 


(*:)  The  following  proposition,  due  to  Rankine,  will  next 
be  proved : 

The  stress  in  every  direction  being  a  thrust,  and  the  great- 
est obliquity  being  given,  it  is  required  to  find  the  ratio  of  two 
conjugate  thrusts  whose  common  obliquity  is  given. 

Let  0  denote  the  greatest  obliquity,  then  we  shall  have  (see 
last  equation  of  §  292) 


0V,  — 


,-+*  = 


Now  let  6  denote  the  common  obliquity  of  two  conjugate 
stresses  whose  intensities  are/  and/,  .  Moreover,  6  <  0,  and 
we  will  consider/,  </.  Then  from  equations  (9)  and  (10)  we 
deduce 


_  4pp*  cosa  B  _  (<rv-<rx  V  . 
"  (/+A)8    ~\<rf  +  oj  ' 

and  combining  this  with  (n)  we  obtain 


RANKINGS  GRAPHICAL   SOLUTION.  889 


COS8  6  -  CQSa  0 


/,  _  cos  6  —  V  cosa  0  —  cosa  0 
~~ 


cos  0  -f  ^cosa  0-cosa  0* 
which  is  the  ratio  shown. 

§  295.  Case  of  any  Stresses  in  Space.  —  In  the  case  of 
stress  which  is  not  all  parallel  to  one  plane,  we  should  find  that 
it  is  always  possible,  no  matter  how  complicated  the  state  of 
stress  in  a  body,  to  find  three  planes  at  right  angles  to  each 
other  on  which  the  stress  is  wholly  normal,  these  being  the 
principal  stresses  ;  and  a  number  of  propositions  follow  analo- 
gous to  those  for  stresses  all  parallel  to  one  plane.  The  discus- 
sions of  these  cases  become  very  complex,  and  will  not  be 
treated  here. 

§  296.  Some  Applications.  —  The  following  are  some  of 
the  practical  cases  which  require  the  theory  of  elasticity  for 
their  solution. 

§  297.  Combined  Twisting  and  Bending.  —  This  is  the 
case  very  generally  in  shafting,  as  the  twist  is  necessary  for 
the  transmission  of  power,  and  the  bending  is  due  to  the  weight 
of  the  pulleys  and  shafting,  and  the  pull  of  the  belts,  this  being 
especially  so  when  there  are  pulleys  elsewhere  than  close  to  the 
hangers  ;  also  in  overhanging  shafts,  in  crank-shafts,  etc. 

Thus  far  we  have  no  tests  of  shafting  under  combined 
twisting  and  bending,  and  therefore  the  methods  used  for 
calculating  such  shafts  vary.  With  many  it  is  the  practice 
to  compute  their  proper  size  from  the  twisting-moment  only, 
but  to  make  up  for  the  bending  by  using  a  large  factor  of 
safety,  the  magnitude  of  this  factor  depending  upon  how  much 


890  APPLIED  MECHANICS. 

the  computer  imagines  the  shaft  will  be  weakened  by  the  par- 
ticular bending  to  which  it  is  subjected. 

With  others  it  is  customary  to  compute  the  deflections, 
under  the  greatest  belt-pulls  that  can  come  upon  it,  by  the 
principles  of  transverse  stress,  without  any  reference  to  the 
torsion,  and  to  so  determine  it  that  the  deflection  computed  in 
this  way  should  not  exceed  ^^V^  or  yg^-g-  of  the  span. 

On  the  other  hand,  Unwin  and  some  others  give  the  for- 
mulae, which  will  be  developed  here  for  combined  twisting  and 
bending,  as  deduced  by  the  theory  of  elasticity.  This  formula, 
has  not,  as  yet,  been  very  extensively  used ;  and  its  constants 
are  taken  from  experiments  on  tension  or  torsion  alone,  and 
not  on  a  combination  of  the  two.  It  is  to  be  hoped  that  we 
may  some  time  have  some  experiments  on  such  a  combination. 
We  will  now  proceed  to  deduce  a  formula  for  the  greatest  in- 
tensity of  the  stress  at  any  point  of  the  shaft. 

For  this  purpose 

Let  Ml  =  bending-moment  at  any  section. 

M2  =  twisting-moment  at  the  same  section. 

7r    =  moment  of  inertia  about  neutral  axis  for  bending. 

72    =  moment  of  inertia  about  axis  of  shaft. 

r     =  distance  from  axis  to  outside  fibre. 

Then,  if  we  denote  by  a-  the  greatest  intensity  of  the  stress 
due  to  bending,  and  by  r  the  greatest  intensity  of  the  stress  due 
to  twisting,  we  have, 


<T   =    ~.  (l)  T   =    -y-  (2) 

For  a  circular  or  hollow  circular  shaft, 
hence 


2  t  N 

--£-•  (3)  r=— .          (4) 


COMBINED    TWISTING  AND  BENDING.  89! 

Then,  at  a  point  at  the  outside  of  the  shaft  in  the  section 
under  consideration,  we  shall  have,  — 
i°.  On  a  plane  normal  to  the  axis, 

(a)  a  normal  stress  <r, 

(b)  a  shearing-stress  T. 

2°.  On  a  plane  in  the  direction  of  the  axis, 

(a)  a  normal  stress  o. 

(b)  a  shearing-stress  T. 

We  thus  have  the  case  solved  in  Example  L,  §  290. 
If,  therefore,  the  greatest  and  least  principal  stresses   be 
denoted  by  or,  and  o-2  respectively,  we  shall  have 


(5) 
(6) 


But,  if  ct  and  e2  denote  the  strains  in  the  directions  of  the 
principal  stresses,  we  have 

77  °"2  °"l  • 

-C€,    =    0-!    —    —  >  £,€2   =    <T2   --  > 

in  m 

Hence,  substituting  for  o-,  and  o-2  their  values,  we  have 


2/tf 


4T2. 


We  then  have,  for  the  greatest  stress  on  any  fibre,  the 
greater  of  the  two  quantities  (7)  and  (8);  and  this  should  not  at 
any  section  of  the  shaft  exceed  the  working-strength  of  the 
material  for  tension. 


APPLIED  MECHANICS. 


The  greater  of  the  two  is  E^  :  hence  we  should  have,  if 
/  =  greatest  stress, 


m  —  i          m  +  i 


2/0 


+  -        -Vcr2  +  47-*  =/. 


If,  now,  we  let  m  =  4,  as  is  commonly  done,  we  have 

fo-  +  IV/CT2  +  4T2  =/  (10) 

this  being  the  formula  given  by  Grashof  and  others  for  com- 
bined twisting  and  bending. 

On  the  other  hand,  Rankine  puts  the  value  of  a-l  in  (5)  equal 
to  /,  and  hence  Rankine's  formula  is 


22  til'f 

This  might  be  derived  from  (9)  by  making  m  =  oo  instead 
of  m  =  4. 

The  formulae  developed  above  are  applicable  to  any  section. 

APPLICATION  TO  CIRCULAR  AND  HOLLOW  CIRCULAR  SHAFTS. 

Substituting  for  a-  and  r  in  (10)  the  values  from  (3)  and  (4), 
we  should  obtain 


which  is  Grashof  s  formula,  and  is  given  by  Unwin  and  others  ; 
and,  substituting  in  (i  i)  instead,  we  should  have 


+  MS)  =/.  (13) 

Equation  (12)  is  equivalent  to  the  following  rule:— 


HOLLOW  CYLINDERS  SUBJECTED    TO  PRESSURE.      89$ 

Calculate  the  shaft  as  though  it  were  subjected  to  a  bending- 
moment 


and  equation  (13)  is  equivalent  to  the  following  rule:  — 

Calculate  the  shaft  as  though  it  were  subjected  to  a  bending- 
moment 


Now,  if,  as  is  usually  the  case,  the  section  where  the  great- 
est bending-moment  acts  is  also  subjected  to  the  greatest 
twisting-moment,  it  will  only  be  necessary  to  put  for  J/i  the 
greatest  bending-moment,  and  for  M2  the  greatest  twisting- 
moment. 

§  298.  Thick  Hollow  Cylinders  subjected  to  a  Uniform 
Normal  Pressure.  —  Let  inside  radius  =  r,  outside  radius  = 
rlt  length  of  portion  under  consideration  —  unity,  intensity  of 
internal  normal  pressure  —  P,  of  external  normal  pressure 

—  P 

—  •*  i* 

i°.  Divide  the  cylinder  into  a  series  of  concentric  rings; 
let  radius  of  any  ring  be  />,  and  thickness  dp,  these  being  the 
dimensions  before  the  pressure  is  applied. 

Let  p  become  p  +  £,  and  dp,  dp  +  d£>  after  the  pressure  is 
applied. 

Then  at  any  point  of  this  ring  we  shall  have,  for  the  strain 
in  the  radial  direction, 

7'  (I) 

dp     . 

and,  since  the  length  of  the  ring  before  the  application  of  the 
pressure  is  27rp,  and  after  is  2ir(p  +  £),  hence  the  strain  in  a 
direction  at  right  angles  to  the  radius  is 


2irp 


8Q4  APPLIED  MECHANICS. 

2°.    Impose,  now,  the  conditions  of  equilibrium  upon   the 
forces  exerted  by  the  rest  of  the  cylinder  upon  the  upper  half- 
ring.     For  this  purpose  let 
p  =  intensity  of  normal  pressure  on  inside  ;  i.e.,  at  dis- 

tance p  from  the  axis. 
f  -|-  dp  —  intensity  of  normal  pressure  on  outside  ;  i.e.,  at  dis- 

tance p  -f-  dp  from  the  axis. 
Then  we  shall  have  for  these  forces,  — 

(a)  Upward  force  due  to  internal  pressure, 


(b)  Downward  force  due  to  external  pressure, 


(c)  Upward  force  at  right  angles  to  radius  acting  at  division 
line  between  the  two  half-rings, 

244.  +  49, 

where  t  =  intensity  of  hoop-tension  per  square  unit;  i.e.,  of 
tension  in  a  circumferential  direction.     Then  we  have 

2(p  +  dp)  (P  +  £  +  dp  +  df)  -  2/(/>  +  i)  -  2t(dp  +  df)  =  o; 

and,  if  this  be  reduced,  and  the  terms 

,     and     *td£ 


be  omitted,  all  of  which  are  very  small  compared  with  the 
remaining  ones,  we  shall  have 


Now,  the  two  stresses  p  and  /  are  principal  stresses,  since 


HOLLOW  CYLINDERS  SUBJECTED    TO  PRESSURE.       895 

there  are  no  shearing-stresses  on  these  planes.    Hence  we  have, 
from  equations  (i)  and  (2),  §  282, 


m 


(5) 


Now  eliminate  /  and  t  between  (3),  (4),  and  (5),  and  obtain 
a  differential  equation  between  p  and  £ 
Proceed  as  follows  :  — 
From  (4)  and  (5), 

m2  —  i\4>        wp/ 
dp  _      Em2    Id2^        i    d£         %  \ 
dp       m2  —  i\dp2       mp  dp       mp2) 

From  (4)  and  (5)  also, 

p  —  t         Em    id&      £\  i 
p       ~  m  +  i  \dp       P/  ft» 

Hence,  substituting  in  (3),  and  reducing,  we  obtain 

P  +  -1  £  -  1          =o  (6) 

ftp         p  dp       p2 

d2^  _       /i  d£  __  £\  _  _      \P/ 
dp2  \pdp       p2)  dp 

Hence,  by  integration, 

2a  being  an  arbitrary  constant,  to  be  determined  from  the  con« 
ditions  of  the  problem. 


896  APPLIED  MECHANICS. 

From  (7)  we  obtain 

<# 

==  20,      or 


Hence,  integrating,  we  have 

tp  =  ap2  +  b,  (8) 

3  being  another  arbitrary  constant. 
From  (8)  we  obtain 

*  =  *P  +  -,  (9> 

P 

which  gives  us,  for  the  two  strains, 

:  II  %—>? 

t  t> 

=  a  +  -2>  (n) 

P  P 

Hence,  substituting  these  values  in  (4)  and  (5),  and  solving 
for/  and  t  successively,  we  obtain 

Em  Em     b  ,     , 

/  =  -  a  --  —  ,  (12) 

m  -  i         m  +  i    2 


i  = a  -\ —  ( i  -2 ) 

m   -    j        r  m   +    !  p2* 

Now,  to  determine  #  and  ^,  we  have  the  conditions,  that, 
when  9  —  r,p  —  Py  and  when  p  =  rlt  p  =  —P,. 
Hence 

P  =     ^**>    j  _     Em      b_     p          Em       _      Em     J>^ 
m  —  i         ^-j-!,^'       I"~W_I         »/  +  i  ^' 


Em        r*  —  r2  m      r*  —  /-2 


HOLLOW  CYLINDERS  SUBJECTED    TO   PRESSURE.      897 

The  greatest  value  of  ty  and  hence  the  greatest  intensity  of 
the  hoop-tension,  occurs  when  P  =  r  ;  and  hence  we  obtain 


Max/=-          r*  _  r2  --  '  ('5) 

this  value  of  t  being  negative  when  there  is  hoop-tension,  be- 
cause the  signs  were  so  chosen  as  to  make  t  positive  when 
denoting  compression. 

If  Pl  =  o,  i.e.,  if  there  is  no  external  pressure,  we  have 


and,  according  to  Professor  Rankine's  method,  we  should  deter- 
mine the  proper  dimensions  by  keeping  max  t  within  the  work- 
ing-strength of  the  material. 

On  the  other  hand,  if  we  decide  that  we  will  keep  the  value 

of  £{-)  within  the  working-strength,  we  shall  find  for  this,  when 


we  make  p  =  r, 


and,  if  m  =  4, 


When  P,  =  o, 

-P&r?  +  f^2) 
-  —^  -  *  --  (19) 

r?  -  r* 

Practical  cases  of  thick,  hollow  cylinders  subjected  to  a  uni 
form  normal  pressure  occur  in  hydraulic  presses  and  in  ord- 
nance. 


898  APPLIED  MECHANICS. 

§  299.  Rankine's  Theory  of  Earthwork. — For  a  mass  of 
earth  bounded  above  by  a  plane  surface,  either  horizontal  or 
sloping,  his  theory  assumes  the  following  proportions  to  be 
true,  viz. :  * 

"  i°.  The  pressure  on  a  plane  parallel  to  the  upper  plane  sur- 
face (which  may  be  called  a  conjugate  plane)  is  vertical,  and 
proportional  to  the  depth. 

"  2°.  The  pressure  on  a  vertical  plane  is  parallel  to  the  up- 
per plane  surface,  and  conjugate  to  the  vertical  pressure. 

"  3°.  The  state  of  stress  at  a  given  depth  is  uniform." 

If  we  let  w  denote  the  weight  of  the  earth  per  unit  of 
volume,  x  the  depth  of  a  given  conjugate  plane  below  the 
surface,  B  the  inclination  of  the  conjugate  plane,  then  is  the 
intensity  of  the  vertical  pressure  on  the  conjugate  plane 

p       zt/       u  ^l^ 

Moreover,  if  0  is  the  angle  of  repose  of  the  earth,  then  is 
0  the  angle  of  greatest  obliquity.  If  now  we  denote  by/  the 
pressure  against  the  vertical  plane,  then  we  have  that 

,  cos  6  —   V  cos2  0  —  cosa"0 
cos  6  -\-  V  cos2  0  —  cos2  0  ' 


,cosfl+  V  cos2  e  -  cos2  0 
cos  0  —    V  cos2  0  ~co<F~<t> '  ™' 

i.e.,  /  may  have  any  value  between  these  two. 

When  the  problem  is  to  determine  the  pressure  exerted  by 
a  mass  of  earth  against  the  vertical  face  ab  of  a  retain  ing- wall, 
we  determine  the  intensity  of  the  pres- 
sure /  against  the  face  (acting  along  cd) 
at  a  depth  x  below  the  upper  surface  ac 
by  means  of  equation  (2)  ;  inasmuch  as 
Rankine  claims  to  prove  by  means  of 
Moseley's  principle  of  least  resistance 
that  the  pressure  against  the  vertical  FlG-  308. 

plane  is  the  lesser  of  the  two  conjugate  pressures. 


RANKINGS    THEORY  OF  EARTHWORK.  899 

Hence,  for  the  entire  pressure  against  the  wall  we  have  a  dis- 
tributed pressure  whose  intensity  is  zero  at  a,  and  which  varies 
uniformly  as  we  go  downwards,  the  direction  of  the  pressure 
being  parallel  to  the  upper  surface  ae,  and  the  point  of  appli- 
cation of  the  resultant  being  at  a  depth  below  a  equal  to  $(ad). 

When  the  upper  surface  ae  is  horizontal,  we  have  cos  6  =  i  ; 
and  hence  (2)  becomes 

i  —  sin  0 


i  -(-  sin  0  t  . 

and  (3)  becomes  /  =  wx  ^  _  ^     .  (5) 

SUPPORTING  POWER  OF  EARTH  FOUNDATION  ACCORDING  TO  RANKINE. 

Let  the  surface  of  the  ground  be  horizontal. 

Then  Rankine  says  that  the  conjugate  pressure  may  be  in- 
creased beyond  the  least  amount  by  the  application  of  some 
external  pressure,  as  the  weight  of  a  building  founded  upon 
the  earth  ;  that,  in  this  case,  the  conjugate  pressure  will  be  the 
least  which  is  consistent  with  the  vertical  pressure  due  to  the 
weight  of  the  building,  and  if  that  conjugate  pressure  does 
not  exceed  the  greatest  conjugate  pressure  consistent  with  the 
weight  of  the  earth  above  the  same  stratum  on  which  the 
building  rests,  the  mass  of  earth  will  be  stable. 

Moreover,  the  greatest  horizontal  pressure  at  the  depth  x, 
consistent  with  stability,  is 

i  -4-  sin  0  ,„. 

p  —  wx  —  —  ^-^  .  (6) 

i  —  sin  0 

The  greatest  vertical  pressure,  consistent  with  this  horizontal 
pressure,  is 

,          i  +  sin  0  /  i  +  sin  0V  ,  . 

p  —p  —  —      ^L  —  'WX\  —  !  —  ^—5)  ;  (7) 

*  i  —  sin  0  V  i  —  sin  0/   ' 

and  this  is  the  greatest  intensity  of  the  pressure  consistent 
with  stability,  of  a  building  founded  on  a  horizontal  stratum 
of  earth  at  a  depth  x,  the  angle  of  repose  being  0. 


900 


APPLIED 


§  300.  Strength  of  Flat  Plates.  —  In  this  regard,  the  for- 
mulae that  will  be  deduced  are  those  of  Professor  Grashof,  the 
reasoning  followed  being  substantially  that  given  by  him  in  his 
"Festigkeitslehre." 

ROUND   PLATES. 

Let  the  curved  line  CA  be  a  meridian  curve  of  the  middle 
layer  of  the  plate  after  it  is 
bent.  Take  the  origin  at  O ; 
let  axis  OZ  be  vertical,  and  axis 
OX  horizontal,  and  let  the  axis 
at  right  angles  to  ZOX  be  O®, 
so  that  #,  xy  and  <£  are  the  co- 
ordinates of  any  point  in  the 
middle  layer  of  the  plate. 

Let  y  denote  the  (vertical) 
distance  of  any  horizontal  layer 
from  the  middle  layer  of  the 
plate. 

Let  R  =  radius  of  curvature 
of  meridian  line  at  any  point 
(x,  z9  «£). 

Let  Rl  =  radius  of  curvature  of  section  of  middle  layer 
normal  to  meridian  line. 

Then  we  should  have,  from  the  differential  calculus, 


(<+(£ 


x  ax 


Hence,  reasoning  in  the  same  way  as  in  the  common  theory  of 
beams,  we  should  have,  for  the  strains  of  the  layer  whose  dis- 


STRENGTff   OF  FLA  T  PLA  TES.  QOI 

tance  from  neutral  layer  is  y  at  point  (x,  </>),  provided  there  is 
no  stress  in  the  plane  of  the  neutral  layer, 

,y  y 

<*  =  ±#    <*  =  ±  j£ 

When  there  is   such  a  stress,  let  the  strains  due  to  that 

stress  be  c^0  and  c$0. 

Then  we  shall  have 

*  g  x 


—  , 


Hence,  substituting  in  (i)  and  (2)  of  §  283,  we  have 


Now  let  us  suppose  the  plate  to  be  subjected,  before  load- 
ing, to  a  uniform  pull  in  its  own  plane,  and  normal  to  its  cir- 
cumference ;  and  let  the  intensity  of  this  pull  be/z.  Then 


and  hence,  we  have, 

-    ) 

(5) 


Therefore,  substituting  in  (3)  and  (4),  and  reducing, 


902  APPLIED   MECHANICS. 

These  equations  express  the  stresses  in  terms  of  the  co- 
ordinates of  the  points. 

Now  impose  the  conditions  of  equilibrium  upon  the  forces 
acting  on  any  half-ring  of  thickness  dx  =  d$. 
These  forces  are  — 

i°.  Force  exerted  upon  it  by  the  outer  part 
of  the  plate, 

\2X<TX  4-    2 d(X(Tx)\dz. 

2°.  Force  exerted  by  the.  inner  part  of  the  plate, 

—  2X(Txdz. 

3°.  Force  exerted  upon  it  by  the  other  half-ring, 


4°.  Force  exerted  by  resistance  to  shear  on  top  and  bottom, 

^  (r  +  dr)  —  r  \  2xdx. 

Hence,  equating  to  zero  the  algebraic  sum  of  these,  and 
reducing,  we  obtain 

dr  __dr       a-^        I  d(xa-x) 
dz  ~~  dy  ~~  x        x      dx 


Now  substitute  for  <rx  and  <r$  their  values,  and  reduce,  and 
we  have 

_- "——i I _ 1  ^o^ 

dy       m*  -  i\<£;3  ^  x  dx*       x2  dx/ 

Integrate  with  regard  to^,  and  we  have,  since  the  quantity/ 
in  brackets  is  not  a  function  of  ^, 


STRENGTH  OF  FLAT  PLATES.  903 

But,  when^/  =  -  (h  being  the  thickness  of  the  plate),  T  =•  o, 
since  there  is  no  shearing-force  at  top  or  bottom  ; 


z        i  d*z        i  dz\ 

£     —.        __    _  I     _  I  _      _        .  _      _     I 

S(m2  —  i)V/*3       x  dy?       x2dxj 


_        m2E(h2  -  ,__ 

S(m2  -  i)      Vfcs        x  ~dx*  ~  &  ~dx)' 

This  gives  us  the  intensity  of  the  shearing-stress  at  any 
point  (*•,  z)  at  distance  y  from  middle  layer  ;  and  this  is  the 
intensity  of  the  shear  at  that  point  between  two  horizontal 
layers,  and  hence  also  along  a  vertical  plane  through  the  point 

(?,  *). 

Now  let  us  take  the  case  of  a  centre  load  P  combined  with 
a  distributed  load  /  per  unit  of  area.  Then  shearing-force  at 
distance  x  from  centre  = 

irx*p  +  P, 

this  tending  to  shear  out  a  circular  piece  of  radius  x.  Hence 
we  must  have  this  balanced  by  the  whole  shearing  resistance 
on  the  surface  subjected  to  shear; 


2-jrx  I    rdy  =  Tt 


Now  substitute  the  value  of  T  from  equation  (10),  integrate, 
and  reduce,  and  we  obtain 


x  dx*       x*  dx 


904  APPLIED  MECHANICS. 


Hence,  for  the  intensity  of  the  shearing-force,  we  have 


This  gives  the  intensity  of  the  shearing-force  at  any  point  of 
the  plate. 

Next,  to  find  its  deflection,  or  the  equation  of  the  meridian 
line,  we  have,  from  (12), 

"  ^  px  +.— 


_         -  i)  p 

P~  ^* 


dx2       x  dx  m2Eh*     (E  2 

d2z       dz  6(m2  -  i)   x*       6(m2  -  i)  P 

.%       X 1 = -p — 5 —  X  log,,  X  +  CX. 

dx2       dx  m2Efa        2  m2Eh*      ir 

But 

^!f  4.  —  --  —(   —\- 
dx2       dx       dx\  dx/ ' 

hence,  integrating,  we  have 

dz_  _         6(m2  —  i)   x4 
dx  ~  m2Eh*        8 

6(^2  _  i)  Px2  6(m2  -  i)  Px2       ex2 


Hence,  dividing  through  by  ^r,  and  integrating, 

—  i)  ^ 
32 


and  this  is  the  meridian  line  of  the  surface,  the  constants  ct  dt 
and  e  being  as  yet  undetermined. 


STRENGTH  OF  FLAT  PLATES.  905 

This  is  as  far  as  we  can  proceed  before  taking  up  special 
cases. 

(a)  Full  Plate.  —  When  the  plate  is  full,  the  slope  becomes 
zero,  for  x  =  o;  therefore  (14)  gives  us 


and  in  this  case  (15)  becomes 
6(m2  -  i)   a* 


,  ,, 
(16) 


32 
6(m2  — 


2J7, 
m2Efc       TT  4 

(a)   Uniformly  Loaded,  no  Centre  Load,  —  P  =  O* 
6(w2  —  i)    ^r4       r*2 


But  when  ;r  =  r,  <sr  =  o  ; 

—  i)  r*       cr* 


— —      32        4 


.  . 

And,  substituting  for  z,  —  and  — ,  their  values  in  (i)  and  (2), 
we  obtain 


and  (13)  gives 


906  APPLIED   MECHANICS. 

(/?)  Supported  all  around.  —  When  x  =.  r,  cr  x  =.  a-^  =  p^  for 
all  values  of  y:  therefore,  from  (6), 

ld*z\        \ldz\ 

m\  -7-  )    +  -(  ~r  )   =  °  > 
\dx*)r       r\dx)r 

/J  W  /T^2  p 

and,  substituting  the  values  of  —  and  —  as  determined   by 

dx         dx* 

differentiating  (18),  we  have,  after  reducing, 


2  m2  Eh* 

Hence  equation  of  meridian  line  is 

-^  ^m  +  x      _       I         _ 

v 


16 
Hence  we  have  maximum  deflection  by  making  x  =  o  ; 

.         ..  3   Q»  - 
"  16 

And,  substituting  in  (19)  and  (20),  we  obtain,  after  reduction, 
m  —  i 


x    v 

'  (24) 

(     j 


But,  in  a  plate  supported  all  around,  /,  =  o  ;  and  then  the 

h 
maximum  value  of  either  one  occurs  when  y  =  -,  and  hence 

I)/..     .      (26) 


STRENGTH  Of  FLAT  PLATES.  907 


On  the   other  hand,  T  becomes  greatest  when  x  =  r  and 
y  =  o.     Hence 

Maxr  =  ?£// 

> 

and,  if  eI  represent  the  maximum  strain  due  to  this  shearing- 
force,  we  have 


RESULTING  FORMULA  FOR  PLATE  SUPPORTED  ALL  ROUND. 


Max  E*  =  3    **-i     3^       i  or        3*jir 

8  *«2  h2  w      /^ 


whichever  is  greatest. 


3   («  -  ')(5«  +  i) 
1  6  ^ 


PLATE  FIXED  AT  ENDS. 

Equation  (17)  applies  to  this  case  also. 

Now,  when  x  =  r,  —  =  o : 
dx 


(28) 


Hence  greatest  deflection  is 


.        3  »ia  -  i  /H 
0       16      /«a      Eh* 


908  APPLIED  MECHANICS. 


and 


When  /,  is  positive  or  zero,  then  Etx  is  maximum  for  x  =  o 
^/  =  -,  and  for  #  =  r,  y  =   --  ;   and  £"^  is   maximum  for 

x  =  O,  >>  =  -  :  and  the  maximum  value  of  E^  is  equal  to  first 

2 


maxmum 


of  Be.     We  have 


Second  max  ^^  =  ^-TLi^  +  1  ^l^l£  ^.  (32) 

/W  4  /«  AJ 

Hence  the  second  is  the  real  maximum. 


RESULTING  FORMULAE  FOR   PLATES  FIXED  AT  THE  ENDS. 
m  —  I  ^          3  W2  —  I  ^2  . 

/,  4- p-  TI/, 

m  4      m2      h2 

.   Z  m*  -  i  pr* 
"  1 6       *rc2      ^^3* 
Por  /,  =  o, 

_,      _          7  m2  —  i  ^2  , 
Max^€0  =  ^ A 

4      w2      ^2 

§  301.  Thickness  of  Plates.  —  Grashof  advises  the  use  of 
3  as  value  of  m.  If  this  be  adopted,  we  should  have,  for  the 
proper  thickness  of  round  plates, 

Supported.  Fixed. 


RECTANGULAR  PLATES. 


909 


where  h  =  thickness,  r  —  radius,  /  =  pressure  per  square  inch> 
and  /  =  working-strength  per  square  inch.  If,  now,  we  use  a 
factor  of  safety  8,  and  use  as  tensile  strength  of  cast-iron  20000,, 
of  wrought-iron,  48000,  and  of  steel  80000,  we  should  have :  — 


Supported. 

Fixed. 

Cast-iron  .     .     . 
Wrought-iron 
Steel     .... 

h  —  0.01825  yorV^ 
h  =  o.oii'jS^oryp 
h  —  0.0091  287^^ 

h  =  o.oi63300?Y/ 
h  =  0.0105410^^ 
h  =  0.0081649?-^ 

§  302.  Rectangular  Plates.  —  Refer  the  plate  to  rectangu- 
lar axes,  as  before,  OZ,  OX,  O<& ;  the  origin  being  at  the  middle 
of  its  middle  layer. 

Let  y  =  distance  of  any  point  in  the  plate  from  the  middle 
layer. 

Let  px  be  the  radius  of  curvature  of  a  normal  section  par- 
allel to  OX  at  the  point  (x,  zt  </>). 

Let  pj,  be  the  radius  of  curvature  of  a  normal  section  par- 
allel to  O®  at  the  point  (x,  z,  <f>). 

Then  we  shall  have,  by  the  principles  of  the  common  theory 
of  beams, 


=  <XQ  ±  A 

p* 


±   A 


where  «^o  and  c^  are  the  strains  of  the  middle  layer  in   the 
directions  OX  and  O<b  respectively. 

Moreover,  from  the  Differential  Calculus,  we  have 


<£f + (I)' 


P* 


z  . 

—  cos2  A.  +  2—  -cosXcos/x,  +  -rrr 


910  APPLIED  MECHANICS. 

where  X  =  angle  between  normal  and  z  axis,  and  //,  =  angle 

between  normal  and  x  axis.     But  -r  and  —  being  the  slopes, 

ax         a<f> 

and  hence  small,  we  shall  have  nearly 

cos  A.  =  i,        cos /A  =  o, 

px  dx2  p<j>  d<fr2 

•••  <*-<*-;£•        '       o>  ) 

_  y/^  (2) 

Hence  (i)  and  (2)  of  §  283  give  us 


mE  mE 

e<f,  )  —  y  -  <  m  --  1 
J 


-       x         <,  —    -       --  --    , 
m*  -  iv  Jm*  -  i(    </*2      "35?* 

w^1  .  mE     (d*z    ,       ^/22) 

0-0  =  -  (ex    +  we,/,  )  —  y  -  \  --  \-  m  -  \  . 
W2_jV^o^  Jm*-i\dx2  d<p\ 

And,  if  o-^,  o-^o,  denote  the  stresses  in  the  middle  layer,  we 
shall  have,  since 

mE  mE 


d2z  ) 

,  (3) 


m2E 


Now,  if  £  and  17  denote  the  increments  in  x  and  <£  respec- 
tively due  to  the  load,  we  shall  have 

£=    f 

Jo 

r 

7  =     /  <    =  <>€^    -  y—  -  /.      _     =    -y 

Jo  d  dx 


RECTANGULAR   PLATES. 


But 


hence 

(5) 

Equations  (3),  (4),  and  (5)  are  the  expressions  giving  the 
stresses  on  two  planes  at  right  angles  to  each  other,  parallel  to 
OX  and  O®  respectively.  Hence  we  have  a  case  of  stress  on 
two  planes  at  right  angles  to  each  other,  and  we  are  to  find  the 
principal  stresses  :  we  thus  have  — 

i°.  Normal  stress  on  x  plane,  cr^ 

2°.  Shearing-stress  on  x  plane,  T^. 

3°.  Normal  stress  on  </>  plane,  o^. 

4°.  Shearing-stress  on  <f>  plane,  T^ 

Hence,  if  we  denote  by  o-I  and  o-2  the  maximum  and  mini- 
mum principal  stress,  we  have  (§  290) 


(6) 


0-2  =  K°-*  +  <r<t>)  -  #(?*  -  *+)*  +  Vxf  ',          (7) 


and  hence,  if  «x  and  e2  denote  the  strains  in  the  directions  of 
the  principal  stresses, 


—    I 


(8) 

*n* 

<TI 


(9) 


912  APPLIED  MECHANICS. 

and  for  the  strain  e3,  parallel  to  OZ,  we  have 

o>  +  gfr 
^€3  =   --  —  •  (10) 

In  order  to  use  (8),  (9),  and  (10),  however,  we  must  know 
<**,  °>>  and  T^  ;  and  for  this  purpose  we  must  know  the  equa- 
tion of  the  middle  layer  after  bending.  For  this  purpose,  apply 
the  equations  (i),  (2),  (3),  of  §  281  to  any  particle  dxd$dz  in 
the  interior  of  the  body.  We  have  then,  X  =  Y  —  Z  =  o. 
Therefore 

d*x      drxz      fa**  _  drxz  _    _/f^f 

dx  "    dy    '  '   d$    ''  dy    ''         \dx  " 


_  t  _         l^±    ,    dfryA 

d$  "   dx    '  '   dy  dy   ''        \d$  "    dx  f 

dvz         dr^z         drxz 

—  --  P  —7-  --  1  --  ;  —   =   O. 

dy         d<$>          dx 

Therefore,  making  use  of  (3),   (4),  and  (5)  with  the  above 
conditions,  we  deduce 

dcrx  n?Ey  Id^-z        i     d^z  \ 

dx  m*  —  \dx*       m  dxd^l 

d<rj>   __  —nPEyld^z         i     d*z  \  f     . 

~d$   "~  m*  —  i\^3  +  m  dtfd^l 


_          mEy      d^z  ,     , 

" 


m 


m  + 
m2Ey  ld*z         d*z   \ 


(     . 


dy 


RECTANGULAR  PLATES.  913 

Hence,  by  integrating  (15)  and  (16),  we  have 
^xz  ==  ~~f    i         n  ~TT  H ; — rrr )  ~i~  *» 


m2Ev2     ld*z          d*z  \ 
~~  2(m2  -  i)\^  +  ~d&d$l  H 


But  when  v  =  -,  r^z  =  rxz  =  o  ; 


ld*z         d*z  \ 

m    -  i\^ 
and1 


\ 


8(m2  —  i)V/<£3 

7  ""  ¥ 


and 

'  j£  +  ^Yy  -  | 

Hence 


w2.g    /^s       _d^_\l&_  _  fc\ 
2  -  i\^       do?d<l>2/\2  "  8/ 


^c 

Now  we  have  <?z  =  /,  where  /  is  the  intensity  of  the  load ; 
therefore  'the  third  equation  gives  us,  on  integrating  between 

the  limits  -  and  —  -, 

k  h 

c 

'dx 


dx 


914  APPLIED  MECHANICS. 


h 
+     m*£    U~  -  ^  H  V:  +  aVrrrr  +  ^  )~ h\  =o 

2 


**  J-^      *  J^-j.JA-2.       '        ^7JA    ~"  ^^2  Z77.7  '  \      // 


and  this  is  the  differential  equation  of  the  surface,  and  should 
be  integrated  in  each  special  case. 


INDEFINITE  PLATES   WHICH   ARE    FIRMLY    HELD    AT    A    SYSTEM    OF 
POINTS   DIVIDING   THEM   INTO   RECTANGULAR  PANELS. 

Let  the  sides  of  the  panels  be  2a  and  2b.  Assume  the 
origin  at  the  middle  of  the  panel,  the  axis  of  x  being  parallel 
to  20,  and  the  axis  of  y  parallel  to  2b.  We  shall  in  this  case 
have  the  following  conditions  ;  viz.,  — 

dz 

(a)  -—  =  o  for  x  =  ±0  and  all  values  of  <£. 

dz 

(b)  —  =  o  for*d>  =  ±£  and  all  values  of  x. 


(c)  2    —  o  when  x  = 

(d)  If  we  develop  the  value  of  z  in  powers  of  x  and  <£,  there 
must  enter  only  even  powers  of  x  and  <£,  since  the  value  of  z 
remains  the  same  when  we  put  —  x  for  x,  or  —  <f>  for  <j>. 

Now,  if  we  write 


-f 
Gx*  4-  .#*4<£2  +  Kx*<p  -I-  Z<#>6  -J-  Mx*y  etc., 


the  above  conditions  will  be  fulfilled  :  — 


RECTANGULAR  PLATES.  915 

i°.  By  making  all  the  co-efficients  after  the  fourth,  each  zero. 
2°.  By  making  D  =  o,  therefore  writing 

0  =  ^4-  Bx*  +  C<f>2  +  Ex*  + 


—  =  2Bx 


Now 


=   O, 

and 

o  =  A  -f  Be?  +  C22  +  ^«4  -|-  ^54 

.-.      B  =  —2JEa2,  C  =  -2F&2, 

.-.     A  =  2Ea*  + 
Hence  the  equation  becomes 

2  =  Ea*  +  7^  -  2^«2^2  -  zFPtf  +  ^"^  -f- 

=  E(a2  - 

—  ==  -4Ex(a? 
dx 

.:      —  =  i2Ex2  — 
dx2 

also 


— 
dx* 


Hence  equation  of  the  middle  layer  is 
=  E(a2  -  x2)2  +  F(P  -  <#>2)2>  where  E  +  /^  =  (w*  ~  ')/      (l8) 


APPLIED  MECHANICS. 


Now,  in  the  case  of  an  ordinary  beam  fixed  at  both  ends, 
and  loaded  uniformly  with  /  Ibs.  per  unit  of  area,  if  b  is  the 
breadth,  we  have  :  — 

i°.  The  points  of  inflection  are  at  a  distance  from  the  middle 

equal  to  ^=,  where  a  is  the  half-span  ;  and 

V3 
2°.  The  bending-moment  at  a  section  at  a  distance  x  from 

the  middle  \*p-(-  -  x*\  when  x  <~  ,  and^r-  -  -}  when 
2\3  /  V/J  z\  3/ 

x  >  —  ;  therefore  the  value  of  #  is  found  from  the  formula 


or 

.  =  j^r(*-3*f+&rr£  -  *-v  when  x 

Etrv'+jJ-*  \        37          £2fcV«   ^o    \3          / 

Vs 

Either  one,  when  integrated,  gives  for  z  the  value 
'  -•• 


Hence  in  the  flat  plate,  if  b  =  o,  the  values  of  E  and  F  must 
be  such  that  the  formula  shall  reduce  to  z  =     ^    (a2  —  ^2)2 

when  3  =  o.     Now,  it  does  reduce  to  s  =  E(a*  —  x2)2.     There- 
fore E  must  be  such  a  function  of  a  and  b,  that,  when  £  =  o,  it 

shall  reduce  to  -£—  .     So  likewise  F  must  be  such  a  function 


of  #  and  £,  that,  when  a  =  o,  it  shall  reduce  to  -—  .     Suppose, 
then,  we  put 


since  these  functions  fulfil  the  above  conditions. 


RECTANGULAR  PLATES. 


Now  we  have 

E  +  F 


r==_ 


Hence,  substituting  for  —  and  -3—  their  values,  and  observ- 
ing that 

€x  is  greatest  for  x  =  ±a,  y  =  ±-, 


is  greatest  for  <£  =  ±^,  ^  =  ±-, 


obtain 


^*.'-5l3rf*      <"> 


2  JJt 

max  (A«A)  =  CTA cr-   ±  2    _  .    .     -rj.        (20) 


Ql8  APPLIED  MECHANICS. 

These  may  be  written  as  follows  : 

.--I/A- 

max  Etx  =  °"x0 o>   ±  2  — 


&* 


^ --(-}" 

i  m\b)    b*          (     . 

—T-*s-  '~WJ-' 


We  have  also,  by  substituting  for  E  and  F  their  values  in 
equation  (18), 


an  --  —t>»  bn  --  —a* 


a* 


•       (23) 


In  these  results  the  exponent  n  is  undetermined,  and  we 
have  no  means  of  determining  it  in  the  general  case.  We  only 
know,  that,  since  the  deflection  must  increase  for  a  decrease  in. 
x  and  <#>,  therefore  we  must  have,  whenever  a  >  b, 

a\n  2  log  m 

-)   <m2 


•   •         '*    \  7      v  * 

Nffl 


This  leaves  the  general  case  indeterminate ;  but  a  common 
practical  case  is  not  subject  to  this  indetermination,  i.e.,  the 
case  when  a  =  bt  for  then 


whatever  the  value  of  n;  and  hence  equations  (21),  (22),  and 
(23)  give 

max  (*„   =        -  I,     ±  5LZJ    ! 


±         _ 


RECTANGULAR  PLATES. 


and 


m2  —  i     a4  ,     x 

max0  =  -          '  T^nA  (27) 


FORMULA  FOR  THE  SHEETS   OF  A  LOCOMOTIVE  FIRE-BOX. 

In  this   case  we   have  a  •=•  b  ;   hence  (24),  (25),  and  (27) 
apply  :  and  if  we  write,  with  Grashof,  m  =  3,  they  become 

max  (Eex)  =<rXo--  <r*o  +  -  ^>,  (28) 


max  (^)  =  o*o  -     <r,o  +  -     /,  (39) 

o  V 

(30) 


Now,  in  the  case  of  the  horizontal  sheets,  cr^  =  o-^  =  o, 
and  we  have 

o        2 

max  (E^)  =  -  -/,  (31) 


(32) 


In  the  case  of  the  vertical  walls,  inasmuch  as  these  have  to 
resist  the  steam-pressure  in  a  vertical  direction,  the  inner  one 
is  called  upon  to  bear  compression,  and  the  outer  tension,  in  a 
vertical  direction.  If  /  is  the  length  of  the  outside  of  the  fire- 
box, and  /,  its  breadth,  we  shall  have  for  the  outer  plate,  taking 
axis  of  x  vertical, 


920  APPLIED   MECHANICS. 

and  for  the  inner  plate,  if  /  and  //  are  corresponding  dimen 
sions  of  inside  of  fire-box, 


And,  by  making  these  substitutions  in  (28),  (29),  and  (30),  we 
obtain  our  formulae. 


RECTANGULAR  PLATE  FIXED  AT  THE  EDGES. 

For  this  case  Grashof  deduces  the  equation  of  the  middle 
layer  as  follows  :  — 

i°.  This  equation  must  be  a  function  of  x  and  <£. 

2°.  If  2a  and  2b  are  the  sides  of  the  plate,  this  function 
must  become 

(a)  When  b  =  oo  for  all  values  of  <£, 


When  a  =  oo  for  all  values  of  x, 


because  the  plate  then  becomes  a  beam  fixed  at  the  ends. 
The  function  that  will  satisfy  these  two  conditions  is 


p 


a*  +  6* 
From  this  he  deduces  for  max  z,  when  x  =  <f>  =  o, 


max,  =     -- 

2  £A*  a* 


RECTANGULAR  PLATES.  Q2I 

From  (i)  he  deduces 


x  .v 


doe-  £fr  a*  + 

d*z  _          2p    (a-  -  x*Y( 


for   - 


these  corresponding  to  the  points  of  inflection  of  a  loaded 
beam  fixed  at  the  ends. 

Hence  (i),  (2),  and  (5)  of  §  300  give 

±--,  (9) 


max  (^)  =  cr,o  -  i^  ±  --     /,  (10) 


8       w         2a*&*     ab  ^  t     . 

max  (ra)  =  ----  /.  in) 

27  *i  4-  i  «4  +  t>*  h2 

At  the  places  where  e^  and  e^  are  greatest, 

TZ   =    O. 

At  the  place  where  rz  is  greatest, 

<r*  =  o-*0,  o-i?  =  <>v 

Hence  it  is  either  (9)  or  (10)  that  gives  us  the  suitable 
formula  to  use  in  any  special  case. 


922  APPLIED  MECHANICS. 


EXAMPLES  OP   THEORY  OP  ELASTICITY. 

1.  It  has  been  sometimes  proposed  to  use  oblique  seams  in  a  boiler- 
shell.     Assume  the  seams  at  an  angle  of  45°  with  the  axis  of  the  boiler,, 
a  pressure  of  100  Ibs.  per  square  inch  of  the  steam,  and  a  diameter  of 
4  feet.     Find  the  tension  per  inch  of  length  of  seam,  and  its  direction. 

2.  Given  a  shaft  carrying  80  HP,  and  running  at  250  revolutions 
per  minute.     Suppose  the  driving-pulley  to  be  at  the   middle   of  the 
length,  this  being  6  feet,  and  given  that  the  ratio  of  the  tension  on  the 
tight  side  of  the  belt  to  that  on  the  loose  side  is  3.75.     Find  the  proper 
size  of  shaft,  assuming  10000  Ibs.  per  square  inch  as  the  working-strength 
of  the  iron. 

3.  What  should  be  the  thickness  of  a  flat  plate  to  bear  150  Ibs. 
pressure  per  square  inch,  and  stayed  at  points  forming  squares  8  inches 
on  a  side,  the  plate  being  of  wrought- iron,  working-strength  10000  Ibs. 
per  square  inch. 

4.  Find   inner  radius   of  a  hydraulic  press  to  bear  1500  Ibs.  per 
square  inch,  given  outer  radius  =  18  inches;  material,  cast-iron;  ten- 
sile strength  20000  Ibs.  per  square  inch. 


INDEX. 


PAGE 

ACCELERATION 75 

Angular  momentum 106 

Arches 779 

conditions  of  stability 800 

correcting  joints 811 

criterion  of  safety 801 

criterion  of  stability 818 

elastic  . 827 

general  remarks 825 

linear 789 

line  of  resistance 795 

line  of  resistance  determined  by  two 

points    . 804 

modes  of  giving  way 790 

Scheffler's  method 805 

true  line  of  resistance 803 

unsymmetrical  arrangement  ....  819 

Atwood's  machine 79 

Axes  of  symmetry  of  plane  figures      .    .  115 

BAR-IRON,  tests  by  Kirkaldy 397 

Barlow,  modulus  of  elasticity  of  wrought- 
iron   395 

Bars  and  shapes  of  wrought-iron,  brands,  423 

Bauschinger,  building-stones 717 

cast-iron  columns  . 375 

cement 733 

repeated  stresses 509 

timber  .    .. 707 

Beams,  assumptions  of  common  theory  .  268 
cross-section  of  equal  strength  .     .     .298 

deflection  of 299 

deflection  with  uniform  bending-mo- 

ment ...    306 

fixed  at  the  ends 312 

load  -not  at  middle 307 

longitudinal  shearing 319 

mode  of  ascertaining  dimensions  .    .  292 
mode  of  ascertaining  stresses    .    .    .291 


PAGE 

Beams,  modulus  of  rupture 293 

moments  of  inertia  of  sections  .    .    .  275 

oak 685 

position  of  neutral  axis 269 

principles  of  common  theory     .    .    .  267 
rectangular,  slope  and  deflection  .    .  311 

resilience  of 306 

shearing- force  and  bending-moment,  272 

slope  and  deflection 301 

slope  and  deflection,  special  cases      .  302 
,  slope  and  deflection  under  working- 
load 310 

.spruce 678 

table  of  deflections  and  slopes  .    .    .  305 
.  table  of  shearing^forces  and  bending- 

raoments 274 

timber,  strength  and  deflection      .    .  671 

timber,  time  tests 689 

uniform  strength  .     .......  296 

variation    of   bending-moment   with 

shearing- force 317 

white-pine 687 

working  strength 293 

wrought-iron,   strength   and    deflec- 
tion   447 

yellowrpine,  strength  and  elasticity,  682 

Beardslee,  effect  of  rest 410 

reduction  in  rolls,  wrought-iron    .     .  409- 

shape  of  specimen    . 408 

tensile  limit 408 

tests  of  wrought-iron 407 

Bending  and  torsion  combined    ....  889 

and  twisting, 338 

Bending-moment 185 

in  beams 272 

Bendingt moments,  graphical  representa- 
tion   387 

Boiler-plate,  brands 422 

Bellman's  truss 219. 

933 


924 


INDEX. 


PAGE 

Bow's  notation 143 

Brands  of  wrought-iron 422 

Breaking-strength  .     .     .    ., 245 

Bridge  columns,  tests  for  Clark,  Reeves 

&  Co 425 

Watertown-arsenal  tests 429 

Bridge-pins 523 

Bridge-trusses    .    .    ,    . 184 

actual  shearing-force '  203 

compound 208 

concentration  of  loads  at  joints    .    .  203 

counterbraces 200 

diagonals 200 

examples 186 

general  formulae 219 

general  remarks 219 

method  of  sections 184 

shearing-force  and  bending-moment  185 
steps  in  determining  stresses  under 

fixed  load 186 

vertical  posts 202 

with  vertical  and  diagonal  bracing    .  198 

Building-stones 714 

Buttress,  stability 793 

CAST-IRON 353 

columns 365 

composition  and  characteristics    .    .  353 
extension  and  compression,  Hodgkin- 

son 360 

-list  of  experimenters 357 

modulus  of  elasticity 361 

modulus  of  elasticity,  Rosset    .    .    .  365 
tensile  and  compressive  strength  .    .  357 

tests  by  Rosset 364 

tests  by  Wade 362 

transverse  strength 381 

'Catenary 784 

transformed 787 

Cement  mortar 722 

Centre  of  gravity 221 

Centre  of  gravity,  examples 226 

of  a  line 225 

of  a  slender  rod 224 

of  flat  plate 223 

of  homogeneous  bodies    .....  223 

of  plane  area 224 

of  solid  bodies 226 

of  symmetrical  bodies 237 

of  system 221 

Pappus's  theorems 234 


PAGE 

Centre  of  percussion 120 

Centre  of  stress 263 

Centres  of  percussion  and  of  oscillation, 

interchangeability 123 

Chain  cable 633 

Chain  or  cord,  loaded 779 

Coefficients  of  expansion 484 

Cold  rolling 495 

Collision 123 

Columns,  cast-iron 365 

Euler's  rules 328 

Gordon's  formulae 325 

Hodgkinson's  rules 331 

oak 652 

strength  of 324 

timber 647 

white-pine 659 

wrought-iron 425 

yellow-pine 651,  664 

Components  of  velocities  of,  and  forces 

acting  on,  a  body 82 

Compound  bridge-trusses 208 

Compression,  direct 253 

of  timber 647 

Compressive  strength  of  cast-iron,  Hodg- 

kinson 357 

of  wrought-iron 423 

Wohler's  experiments 254 

Continuous  girders 743 

distributed  and  concentrated  loads    .  770 

examples  for  practice 778 

loads  concentrated 763 

loads  distributed 744 

Contraction  of  area,  Kirkaldy     ....  402 

Cord  or  chain,  loaded 779 

Cord  with  load    uniformly    distributed 

horizontally 782 

Counterbraces 200 

Couple,  composition  of,  in  inclined  planes,   59 

Couples,  composition  of 57 

effect  on  rigid  body       55 

effect  when  forces  are  inclined  to  rod    54 
measure  of  rotatory  effect     ....    53 

moment  of         53 

effect  on  rigid  rod 51 

representation  by  a  line 59 

resultant  with  single  force    .    .    .    .    61 

Crystalline  fracture,  Kirkaldy    ....  402 

Crystallization  of  iron  and  steel       .    .    .  475 

Cylinders,  thick  hollow,  strength  of   .    .  893 

thin  hollow .  251 


INDEX. 


92$ 


PAGE 

DEFLECTION  of  beams 299,  301 

Domes 843 

Dynamics 75 

ECCENTRIC  load  on  columns    .    .  371, 654,  657 
Elasticity,  modulus  for  timber  beams  .    .  696 

modulus,  cast-iron 361 

modulus  for  use  with  spruce  beams  .  677 

modulus  of 240 

modulus,  Rosset 365 

modulus,    wrought-iron,     Hodgkin- 

son 393 

modulus,  wrought-iron,  Barlow    .    .  395 
special  modulus,  wrought-iron,  Ros- 
set      418 

tensile  modulus,  wrought-iron,  Water- 
town  Arsenal 414 

theory  of 852 

Ely,  rules  for  strength  of  timber.posts    .  669 

Energy 78 

Equilibrium  curves 779 

Euler's  rules  for  columns 328 

Expansions,  coefficients 484 

Eye-bars,  steel,  Hill 465 

steel,  Watertown  Arsenal     ....  467 
wrought-iron 419 

FACTOR  of  safety  for  iron  and  steel     .    .  497 

timber 670 

Fibrous  fracture,  Kirkaldy 402 

Fink's  truss 217 

Floors,  timber 699 

Force 3 

and  momentum,  relation  between     .    n 
applied  at  centre  of  gravity  of  rigid 

rod 51 

applied  to  rigid  rod,  not  at  centre     .     46 

centrifugal .    .     .    81 

centrifugal,  of  solid  body     ....    85 

characteristics  of 16 

criticism  of  definition 6 

definition  of 8 

deviating 82 

external    9 

intensity  of,  distributed 40 

measure  of 5,  9,  76 

moment  of 30 

relativity  of 9 

resultant  of,  distributed 40 

single,  at  centre  of  rigid  rod      ...    43 
Forces,  centre  of  system  of  parallel     .    .    38 


Forces,  composition  of 21,  23- 

composition  of  parallel  .  .  .  .  37, 62 
composition  of  parallel,  in  a  plane  .  36 

composition  of  two 30 

co-ordinates  of  centre  of  parallel  .    .    39 

decomposition  of I9 

distributed 39 

effect  of  pair  on  rigid  rod      ....    50 

equilibrium  of 28 

equilibrium  of,  in  a  plane      ....    69- 

equilibrium  of,  in  space 73 

equilibrium  of  parallel 37- 

equilibrium  of  three  parallel ....  31 
moment,  causing  rotation  ....  49 
normal  and  tangential  components  .  8a 

parallelogram  of 16. 

polygon  of 23 

resultant  of  any  number  of  parallel  .  35 
resultant  of,  in  a  plane  .....  66 

resultant  of,  in  space 7o. 

resultant  of  two  parallel 33 

statical  measure  of i» 

triangle  of 19. 

Foundry  iron 354 

Frame,  triangular 141 

isosceles  triangular 143 

polygonal i45 

Frames  of  two  bars 138 

Frames,  stability 140 

Framing- joints 698 

Friction  of  blocks jgi 

Funicular  polygon 147 

G,  relation  to  E 869 

G,  value  of 870 

Girder,  greatest  stresses 193 

Girders,  continuous 743 

distributed  and  concentrated  loads  .  770 

examples  for  practice 778 

loads  concentrated 763 

Gordon's  formulae 325 

Gordon's  formula,  applicability  to  bridge 

columns 425 

modified  forms 427 

Grade,  effect  on  tractive  force    .     .    .     .  100 
Gravity,  centre  of 42,221 

HALF-LATTICE    girder  :    travelling-load, 

greatest 192 

Hammer-beam  truss 176 

wind  pressure 179* 


926 


INDEX. 


PAGE 

Harmonic  motion 102 

Hill,  steel  eye-bars      ........  465 

steel  plates .  473 

Hodgkinson,   tensile  strength  and  elas- 
ticity of  wrought-iron 393 

Hodgkinson's rules  for  columns.     .    .     .331 

tests  of  cast-iron  columns 365 

tests  on  tension  and  compression  of 
cast-iron 358 

Hooks,  strength  of 322 

IMPACT 123 

central 124 

co-efficient  of  restitution 125 

co-efficient  of  restitution  by  experi- 
ment  132 

elastic  .     .   • 127 

imperfectly  elastic 129 

inelastic 126 

oblique 134 

of  revolving  bodies 136 

special  cases  of  elastic 128 

Impact,  special  cases  of  imperfect  elas- 
ticity  132 

special  cases  of  inelastic 128 

velocity  at  greatest  compression    .     .  125 

KIRKALDY,  riveted  joints 630 

tests  of  bar  iron 397 

tests  of  iron  plate 400 

tests  of  rivet  iron  . 399 

tests  of  wrought-iron 396 

Kirkaldy's  conclusions 401 

LAUNHARDT'S  formula 248,  501 

Load,  sudden  application  of 246 

Longitudinal  shearing  of  beams      .    .    .319 

MASS,  measure  of 10 

unit  of 13 

Materials,  strength  of 240 

Metals  and  alloys  other  than  iron  and 

steel 486 

Modulus  of  elasticity      . 240 

approximate  values 245 

Modulus  of  rapture,  for  beams   ....  293 

Moeller,  cast-iron  columns 379 

Moment  of  deviation 108 

Moment  of  inertia 106 

Moments  of  inertia  about  different  axes,  113 
components  of 117 


PAGE 
Moments  of  inertia,  equal  values  of    .    .116 

examples. n8 

of  Phoenix  eolum-ns 286 

of  plane  figures  about  parallel  axes  .  no 

of  plane  surface 107 

of  sections 275 

of  solids  around  parallel  axes   .    .    .117 

polar,  of  plane  figures m 

principal 114 

Momentum n 

Mortar  (cement) 722 

Motion  and  rest 2 

Motion,  Newton's  first  law  of     ".    .    .    .  4,  9 

Newton's  second  law  of 13 

on  curved  line 95 

on  inclined  plane 92 

relativity  of ,    .    .      i 

Motion,  under  influence  of  gravity     .    .    87 

uniform 76 

uniformly  varying 76 

uniformly  varying  rectilinear    ...    87 

Motions,  parallelogram  of 15 

polygon  of 15 

m,  value  of 870 

NAILS  in  one  pound 151 

Neutral  axis  of  beams 269 

Notation,  Bow's 143 

OAK  columns 652 

PAPPUS'S  theorems 234 

Pendulum,  cycloidal 99 

simple  circular 97 

Pig-iron .353 

Plate-iron,  tests  by  Kirkaldy  .....  400 
Plates,  flat,  strength  of 

steel,  Hill 473 

thickness  of 908 

Polygonal  frame 145 

Projectile,  unresisted 89 

Punching  and  drilling  plates 535 

effect  on  steel,  Hill 474 

RADIUS  of  gyration 107 

Rankine,  strength  of  timber 642 

Reduction  in  rolls,  Beardslee,  wrought- 
iron  409 

Resilience  of  a  beam 306 

of  tie-bar 247 

Resistance,  line  of 795 


INDEX. 


927 


PAGE 

Resistance,  line  of  maximum  and  mini- 
mum       799 

to  direct  compression 253 

to  shearing 256 

true  line  of 803 

Rest,  effect,  Beardslee 410 

Rigid  bodies,xrectilinear  transference  of 

force 29 

rotation  of 105 

statics  of 29 

Riveted  joints 526 

Riveted  joints,  Kirkaldy 630 

proportions  given  by  Box  ....  529 
proportions  given  by  Unwin  .  .  .531 
proportions  given  by  Wilson  .  .  .528 

punching  and  drilling 535 

Watertown-arsenal  tests 548 

Rivet-iron,  tests  by  Kirkaldy 399 

Rodman,  strength  of  timber 645 

Roofs,  estimation  of  load 163 

weight  of  materials 151 

Roof-trusses 138,  166 

determination  of  stresses  .    .    .    150,  165 

distribution  of  load 163 

examples 183 

general  remarks 172 

with  loads  at  lower  joints     ....  171 

Rope,  wire 635 

Rosset,  special  modulus  of  elasticity  of 

wrought-iron 418 

tests  of  cast-iron 364 

Rupture,  modulus  for  beams 293 

SCHEFFLER'S  method  with  arches     .     .    .  805 

mode  of  correcting  joints 811 

Scissor-beam  truss 179 

without  horizontal  tie 180 

Seasoned  columns 653 

Seasoning,  effect  on  timber 712 

Semi-girder,  greatest  stresses 193 

Shafting,  strength  of 333,  515 

Shafts,  transverse  deflection 337 

under  combined  torsion  and  bend- 


ing 


Shape  of  specimen,  Beardslee  ....  408 
Kirkaldy 4°5 

Shearing-force 185 

actual  for  bridge-trusses 203 

in  beams 272 

Shearing,  resistance  to 256 

Shearing-strength  of  iron  and  steel     .     .  514 


PAGE 

Shearing  of  timber 7n 

of  timber  beams 696 

Slating,  weight 151 

Slope  of  beams 301 

Slotted  cross-head 102 

Snow,  weight  of 151 

Springs 338 

torsional  strength 339 

transverse  strength 343 

Spruce  beams 678 

Stability  of  a  buttress 793 

of  an  arch goo 

of  position 793 

Steel,  composition,  kinds,  and  character- 


istics   452 

crystallization 475 

effect  of  cold  rolling 495 

effects  of  temperature 483 

eye-bars,  Hill 465 

eye-bars,  Watertown  Arsenal  .    .    .467 

list  of  experimenters 455 

plates,  Hill 473 

tensile  strength  and  elasticity,  Water- 
town  tests 459 

torsional  strength 515 

transverse  strength 447 

wire 634 

Stone 714 

Strain 240 

resultant 857 

Strains 852 

in  terms  of  distortions 855 

relation  to  stresses 865 

Strength,  breaking  and  working    .     .    .245 

of  columns 425 

of  hooks 322 

of  materials 240 

of  materials,  general  remarks    .    .    .350 
of  shafting 333 

Stress     .......  240 

centre  of 263 

compound 861 

converse  of  ellipse 884 

ellipse 881 

graphical  representation 261 

intensity  of 260 

relation  to  strains 865 

simple 860 

tangential 861 

uniform 264 

uniformly  varying 265 


928 


INDEX. 


Stress,    uniformly    varying,    amounting 

to  a  statical  couple 266 

Stresses 859 

Stresses,  composition  of 872 

conjugate 872 

equilibrium  of 865 

greatest,  in  girder 193 

in  roof-trusses,  determination  of  .  .  150 
mode  of  ascertaining,  in  a  beam  .  .  291 
parallel  to  a  plane,  composition  .  ,  875 

principal 877 

principal,  determination  of    ....  878 

Stretching  and  tearing 242 

Strut -138 

Struts,  short 323 

Suspension-rod  of  uniform  strength    .    .  250 

TEMPERATURE,  effect  on  iron  and  steel    .  483 

Tensile  limit,  Beardslee 378 

Tension  of  timber 646 

Tests  of  boiler-plate 421 

Theory  of  elasticity 852 

Tie 148 

Timber    beams,   immediate  modulus  of 

elasticity 696 

beams,    modulus     of    elasticity    for 

use 696 

beams,  strength  and  deflection  .    .    .  671 

beams,  time  tests 689 

columns 647 

columns  loaded  eccentrically    .    654,  657 

columns,  seasoned 653 

compression      .........  647 

effect  of  seasoning 712 

factor  of  safety 670 

floors 699 

framing-joints 698 

general  remarks 712 

list  of  experimenters 641 

longitudinal  shearing  in  beams  .  .  696 
posts,  rules  for  strength,  Ely  .  .  .  669 

shearing 711 

strength  as  given  by  Rankine  .  .  .  642 
strength  as  given  by  Rodman  .  .  .  645 
strength,  general  remarks  ....  640 

tension 646 

tests  by  Bauschmger 707 

transverse  strength 671 

Time  of  descent  down  a  curve    ....    96 

Time  tests  on  timber  beams 689 

Torsional  strength  of  iron  and  steel    .    .515 


PAGK- 

Torsional  strength  of  springs  .  .  r  ,  .  335 
Torsion  and  bending  combined  ....  889 
Translation  and  rotation  combined  .  .  44. 
Transverse  deflection  of  sh?f*s  ....  337 
Transverse  strength  of  cast-  .'.\  on  .  .  .381 

springs 343 

steel 447 

timber 671 

wrought-iron 447 

Travelling-load 192 

Triangular  frame 141 

Triangular  truss,  wind  pressure  ....  147 

Truss,  Bellman's 219 

Fink's 217 

hammer-beam 176 

scissor-beam 179 

triangular,  wind  pressure 147 

Trusses,  bridge 184 

methods  for  determining  stresses  .    .  140 

roof 138,  166 

roof,  determination  of  stresses  .    .     .  150 
Twisting  and  bending  combined     .     .     .  338 

VELOCITY 2,  75 

WADE,  tests  of  cast-iron 362 

Watertown  Arsenal,  steel  eye-bars    .     .  467 

tensile  strength  and  elasticity  of  steel  459 

tensile    strength     and    elasticity    of 
wrought-iron 414 

tests  of  bridge  columns 429 

tests  of  riveted  joints 548 

Weight  of  materials  for  roofs 151 

of  snow 151 

Weyrauch's  formula 255,  504 

White-pine  beams 687. 

columns 659 

Wind  pressure 152 

triangular  truss     ........  147 

Wire,  iron  and  steel 634 

rope 635 

Wohler's  experiments      . 499 

on  compressive  strength 254 

on  tensile  strength 247 

results 506 

Working-strength 245 

of  beams  .     .     .     .    , 293 

Work,  mechanical 77 

under  oblique  force 104. 

unit   of 77 


INDEX. 


929 


PAGE 

Wrought-iron  beams,  strength  and  de- 
flection       447 

Wrought-iron  characteristics 391 

columns,  tests,  full-size 425 

compressive  strength 423 

crystallization 475 

effect  of  temperature 483 

eye-bars 419 

list  of  experimenters 392 

tensile  and  compressive  strength  .    .  392 


PAGE 

Wrought-iron,  tensile  strength  and  elas- 
ticity, Hodgkinson 393 

tensile  tests  at  Watertown  Arsenal    ,  414 

tests  by  Beardslee 407 

tests  by  Kirkaldy 396 

Transverse  strength ,  447 

YELLOW-PINE   beams,  strength  and  elas- 
ticity       682 

columns,  tables 651,  664 


UNIVERSITY  OF  CALIFORNIA  LIBRARY 
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